Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Miscellaneous Exercise on Chapter 7

Integration is the inverse process of differentiation. In the differential calculus, we are given a function, and we have to find the derivative or differential of this function, but in the integral calculus, we are to find a function whose differential is given. Thus, integration is a process which is the inverse of differentiation.

Some Important formulas of integration are given below:

In this article, we will discuss all the solutions for Miscellaneous Exercise of Chapter 7 of Class 12 NCERT.

Question 1:[Tex]\frac{1}{x-x^3}[/Tex]

Answer:

[Tex]\frac{1}{x-x^3} = \frac{1}{x(1-x^2)}=\frac{1}{x(1-x)(1+x)}[/Tex]

[Tex]let\,\,\frac{1}{x(1-x)(1+x)} = \frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}\,\,\,\,……….(i)[/Tex]

[Tex]=> 1 = A(1-x^2)+Bx(1+x)+Cx(1-x) \\=> 1 = A-Ax^2+Bx+Bx^2+Cx-Cx^2[/Tex]

Equating the coefficient of x² ,x and constant term, we obtain

-A+B-C = 0

B+C = 0

A = 1

On solving this equation, we obtain

[Tex]A = \frac{1}{2},and\,\, C = -\frac{1}{2}[/Tex]

From equation (1),we obtain

[Tex]\frac{1}{x(1-x)(1+x)}=\frac{1}{x}+\frac{1}{2(1-x)}-\frac{1}{2(1+x)}[/Tex]

[Tex]=>\frac{1}{x(1-x)(1+x)}dx=\int \frac{1}{x}dx +\int \frac{1}{2(1-x)} dx-\int \frac{1}{2(1+x)}dx[/Tex]

=log|x|-[Tex]\frac{1}{2}[/Tex]log|(1-x)|-[Tex]\frac{1}{2}[/Tex]log|(1+x)|

=log|x| – log| [Tex](1-x)^\frac{1}{2}[/Tex]|-log| [Tex](1+x)^\frac{1}{2}[/Tex] |

=log| [Tex]\frac{x}{(1-x)^\frac{1}{2}(1+x)^\frac{1}{2}}[/Tex] | + C

= [Tex]\frac{1}{2}log|\frac{x^2}{1-x^2}|+C[/Tex]

Question 2: [Tex]\frac{1}{\sqrt{x+a}+\sqrt{x+b}}[/Tex]

Answer:

[Tex]\frac{1}{\sqrt{x+a}+\sqrt{x+b}}[/Tex] [Tex]=\frac{1}{\sqrt{x+a}+\sqrt{x+b}}*\frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}[/Tex]

[Tex]=\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}[/Tex]

[Tex] =\frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}[/Tex]

[Tex]=> \int \frac{1}{\sqrt{x+a}+\sqrt{x+b}}dx[/Tex]=[Tex]\frac{1}{a-b}\int (\sqrt{x+a}-\sqrt{x+b})dx[/Tex]

[Tex]=\frac{1}{a-b}[ \,\frac{(x+a)^\frac{3}{2}}{\frac{3}{2}}-\frac{(x+b)^\frac{3}{2}}{\frac{3}{2}}] \,[/Tex]

[Tex]=\frac{2}{3(a-b)}[ \,(x+a)^\frac{3}{2}-(x+b)^\frac{3}{2} ] \,+C[/Tex]

Question 3:[Tex] \frac{1}{x \sqrt{ax – x^2}} \, [/Tex]

Answer:

[Tex]\int \frac{1}{x \sqrt{ax – x^2}} \, dx\\ Let \, x = \frac{a}{t} \implies dx = -\frac{a}{t^2} dt\\ \implies \int \frac{1}{x \sqrt{ax – x^2}} \, dx = \int \frac{1}{\frac{a}{t} \sqrt{a \cdot \frac{a}{t} – \left(\frac{a}{t}\right)^2}} \left(-\frac{a}{t^2} dt\right) \\= -\int \frac{1}{a/t \cdot \sqrt{\frac{a}{t} \left(a – \frac{a}{t}\right)}} \cdot \frac{a}{t^2} dt \\= -\int \frac{1}{at} \cdot \frac{1}{\sqrt{\frac{1}{t} – \frac{1}{t^2}}} dt \\= -\frac{1}{a} \int \frac{1}{\sqrt{t^2 – t}} dt \\= -\frac{1}{a} \int \frac{1}{\sqrt{t – 1}} dt \\= -\frac{1}{a} \left[ 2\sqrt{t – 1} \right] + C \\= -\frac{1}{a} \left[ 2\sqrt{\frac{a}{x} – 1} \right] + C \\= -\frac{2}{a} \left( \sqrt{\frac{a – x}{x}} \right) + C \\= -\frac{2}{a} \left( \sqrt{\frac{a – x}{x}} \right) + C [/Tex]

Question 4:[Tex] \frac{1}{x^2 \left( x^4 + 1 \right)^{\frac{3}{4}}} \, [/Tex]

Answer:

[Tex]\int \frac{1}{x^2 \left( x^4 + 1 \right)^{\frac{3}{4}}} \, dx \\Multiplying\,\, and\,\, dividing\,\, by \,\,\, x^{-3} \, we\,\, obtain \frac{x^{-3}}{x^2 \cdot x^{-3} \left( x^4 + 1 \right)^{\frac{3}{4}}} = \frac{x^{-3} \left( x^4 + 1 \right)^{-\frac{3}{4}}}{x^2 \cdot x^{-3}} \\= \frac{\left( x^4 + 1 \right)^{-\frac{3}{4}}}{x^5 \cdot \left( x^4 \right)^{\frac{3}{4}}} \\= \frac{\left( x^4 + 1 \right)^{-\frac{3}{4}}}{x^5 \cdot \left( x^4 \right)^{\frac{3}{4}}} \\= \frac{1}{x^5 \left( \frac{x^4 + 1}{x^4} \right)^{\frac{3}{4}}} \\= \frac{1}{x^5 \left( 1 + \frac{1}{x^4} \right)^{\frac{3}{4}}} \\Let \frac{1}{x^4} = t \implies -\frac{4}{x^5} dx = dt \implies \frac{1}{x^5} dx = -\frac{dt}{4} \\ \therefore \int \frac{1}{x^2 \left( x^4 + 1 \right)^{\frac{3}{4}}} \, dx \\= \int \frac{1}{x^5 \left( 1 + \frac{1}{x^4} \right)^{\frac{3}{4}}} \, dx \\= \int \frac{1}{x^5 \left( 1 + t \right)^{\frac{3}{4}}} \, dx \\= -\frac{1}{4} \int (1 + t)^{-\frac{3}{4}} dt \\= -\frac{1}{4} \left[ (1 + t)^{\frac{1}{4}} \cdot \frac{4}{1} \right] + C \\= -\left( 1 + t \right)^{\frac{1}{4}} + C \\= -\left( 1 + \frac{1}{x^4} \right)^{\frac{1}{4}} + C [/Tex]

Question 5:[Tex] \frac{1}{x^2 + x^3} \, \\\\Hint: \frac{1}{x^2 + x^3} = \frac{1}{x^3 \left( \frac{1}{x} + 1 \right)} = \frac{1}{x^3 \left( 1 + \frac{1}{x^6} \right)} \quad \text{Put } x = t^6 [/Tex]

Answer:

[Tex] \int \frac{1}{x^2 + x^3} \, dx = \int \frac{1}{x^3 \left(1 + x^6\right)} \, dx \\\text{Let } x = t^6 \implies dx = 6t^5 dt \\\therefore \int \frac{1}{x^2 + x^3} \, dx \\= \int \frac{1}{t^{12} \left(1 + t^{12}\right)} \cdot 6t^5 \, dt \\= \int \frac{6t^5}{t^{12} (1 + t^{12})} \, dt\\ = \int \frac{6}{t^7 (1 + t^{12})} \, dt = 6 \int \frac{t^3}{1 + t} \, dt \\On\ dividing, we \,obtain\\ \int \frac{1}{x^2 + x^3} \, dx = 6 \int \left( \frac{t^2 – t + 1}{1 + t} \right) \, dt \\= 6 \left[ \frac{t^3}{3} – \frac{t^2}{2} + t – \log |1 + t| \right] \\= 2x^{\frac{1}{2}} – 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} – 6 \log \left( 1 + x^{\frac{1}{6}} \right) + C \\= 2\sqrt{x} – 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} – 6 \log \left( 1 + x^{\frac{1}{6}} \right) + C [/Tex]

Question 6:[Tex]\frac{5x}{(x+1)(x^2+9)} [/Tex]

Answer:

[Tex]Let \frac{5x}{(x+1)(x^2+9)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+9} \quad \cdots (1) \\\Rightarrow 5x = A(x^2 + 9) + (Bx + C)(x + 1) \\ \Rightarrow 5x = Ax^2 + 9A + Bx^2 + Bx + Cx + C \\ \Rightarrow 5x = (A + B)x^2 + (B + C)x + 9A + C \\ Equating \,the\, coefficients\, of (x^2, x), and\, the\, constant \,term,\,\\ we\, obtain: \begin{cases} A + B = 0 \\ B + C = 5 \\ 9A + C = 0 \end{cases} \\On \,solving \,these\, equations, we \,obtain: \\ A = -1, \quad B = 1, \quad \text{and} \quad C = \frac{9}{2} \\From\, equation (1), we \,obtain: \\ \frac{5x}{(x+1)(x^2+9)} = \frac{-1}{2(x+1)} + \frac{x + \frac{9}{2}}{x^2 + 9} \\ \int \frac{5x}{(x+1)(x^2+9)} \, dx = \int \left( \frac{-1}{2(x+1)} + \frac{x + 9/2}{x^2 + 9} \right) \, dx \\= \int \frac{-1}{2(x+1)} \, dx + \int \frac{x}{x^2 + 9} \, dx + \int \frac{9/2}{x^2 + 9} \, dx \\= -\frac{1}{2} \log|x+1| + \frac{1}{2} \int \frac{2x}{x^2+9} \, dx + \frac{9}{2} \int \frac{1}{x^2+9} \, dx \\= -\frac{1}{2} \log|x+1| + \frac{1}{4} \log|x^2+9| + \frac{9}{2} \cdot \frac{1}{3} \tan^{-1} \frac{x}{3} \\= -\frac{1}{2} \log|x+1| + \frac{1}{4} \log|x^2+9| + \frac{3}{2} \tan^{-1} \frac{x}{3} + C [/Tex]

Question 7: [Tex] \frac{\sin x}{\sin (x – a)} \, [/Tex]

Answer:

[Tex] \frac{\sin x}{\sin (x – a)} \, dx \\Let \,(x – a) = t \Rightarrow dx = dt\\ \int \frac{\sin x}{\sin (x – a)} \, dx = \int \frac{\sin (t + a)}{\sin t} \, dt \\= \int \frac{\sin t \cos a + \cos t \sin a}{\sin t} \, dt \\= \int \left(\cos a + \cot t \sin a\right) \, dt \\ = t \cos a + \sin a \log |\sin t| + C_1 \\= (x – a) \cos a + \sin a \log |\sin (x – a)| + C_1 \\= x \cos a + \sin a \log |\sin (x – a)| – a \cos a + C_1 \\= \sin a \log |\sin (x – a)| + x \cos a + C [/Tex]

Question 8: [Tex]\frac{e^{5 \log x} – e^{4 \log x}}{e^{3 \log x} – e^{2 \log x}}[/Tex]

Answer:

[Tex]\frac{e^{5 \log x} – e^{4 \log x}}{e^{3 \log x} – e^{2 \log x}} \\\frac{e^{5 \log x} – e^{4 \log x}}{e^{3 \log x} – e^{2 \log x}} = \frac{e^{4 \log x} \left(e^{\log x} – 1\right)}{e^{2 \log x} \left(e^{\log x} – 1\right)} \\= e^{2 \log x} \\= e^{\log x^2} \\= x^2\\ \therefore \int \frac{e^{5 \log x} – e^{4 \log x}}{e^{3 \log x} – e^{2 \log x}} \, dx \\= \int x^2 \, dx = \frac{x^3}{3} + C [/Tex]

Question 9: [Tex]\frac{\cos x}{\sqrt{4 – \sin^2 x}}[/Tex]

Answer:

[Tex]\\\text{Let } \sin x = t\\ \implies \cos x \, dx = dt \\\implies \int \frac{\cos x}{\sqrt{4 – \sin^2 x}} \, dx = \int \frac{dt}{\sqrt{(2)^2 – (t)^2}} \\= \sin^{-1} \left( \frac{t}{2} \right) + C \\= \sin^{-1} \left( \frac{\sin x}{2} \right) + C [/Tex]

Question 10: [Tex]\frac{\sin^8 x – \cos^8 x}{1 – 2 \sin^2 x \cos^2 x}[/Tex]

Answer:

[Tex]\\\frac{\sin^8 x – \cos^8 x}{1 – 2 \sin^2 x \cos^2 x} \\= \frac{(\sin^4 x + \cos^4 x)(\sin^4 x – \cos^4 x)}{\sin^2 x + \cos^2 x – \sin^2 x \cos^2 x – \sin^2 x \cos^2 x} \\= \frac{(\sin^4 x + \cos^4 x)(\sin^2 x + \cos^2 x)(\sin^2 x – \cos^2 x)}{\sin^2 x – \sin^2 x \cos^2 x + (\cos^2 x – \sin^2 x \cos^2 x)} \\= \frac{(\sin^4 x + \cos^4 x)(\sin^2 x – \cos^2 x)}{\sin^2 x(1 – \cos^2 x) + \cos^2 x(1 – \sin^2 x)} \\= \frac{(\sin^4 x + \cos^4 x)(\sin^2 x – \cos^2 x)}{\sin^2 x – \sin^4 x + \cos^2 x – \cos^4 x} \\= – \cos 2x \\\therefore \int \frac{\sin^8 x – \cos^8 x}{1 – 2 \sin^2 x \cos^2 x} \, dx \\= \int -\cos 2x \, dx \\= – \frac{\sin 2x}{2} + C [/Tex]

Question 11: [Tex]\frac{1}{\cos(x + a) \cos(x + b)}[/Tex]

Answer:

[Tex] \frac{1}{\cos(x + a) \cos(x + b)} \\\text{Multiplying and dividing by } \sin(a – b), \text{ we obtain} \\\frac{1}{\sin(a – b)} \left[ \frac{\sin(a – b)}{\cos(x + a) \cos(x + b)} \right] \\= \frac{1}{\sin(a – b)} \left[ \frac{\sin[(x + a) – (x + b)]}{\cos(x + a) \cos(x + b)} \right] \\= \frac{1}{\sin(a – b)} \left[ \frac{\sin(x + a) \cos(x + b) – \cos(x + a) \sin(x + b)}{\cos(x + a) \cos(x + b)} \right] \\= \frac{1}{\sin(a – b)} \left[ \frac{\sin(x + a)}{\cos(x + a)} – \frac{\sin(x + b)}{\cos(x + b)} \right] \\= \frac{1}{\sin(a – b)} \left[ \tan(x + a) – \tan(x + b) \right] \\\int \frac{1}{\cos(x + a) \cos(x + b)} \, dx = \frac{1}{\sin(a – b)} \left[ \int \left( \tan(x + a) – \tan(x + b) \right) dx \right] \\= \frac{1}{\sin(a – b)} \left[ -\log|\cos(x + a)| + \log|\cos(x + b)| \right] + C \\= \frac{1}{\sin(a – b)} \log \left| \frac{\cos(x + b)}{\cos(x + a)} \right| + C [/Tex]

Question 12: [Tex] \frac{x^3}{\sqrt{1-x^8}} [/Tex]

Answer:

[Tex]\frac{x^3}{\sqrt{1-x^8}}\\ Let\ x^4 = t\\ \implies 4x^3 dx = dt\\ \Rightarrow \int \frac{x^3}{\sqrt{1-x^8}} dx \\= \frac{1}{4} \int \frac{dt}{\sqrt{1-t^2}}\\ = \frac{1}{4} sin^{-1} t + C = \frac{1}{4} sin^{-1} (x^4) + C[/Tex]

Question 13: [Tex] \frac{e^x}{(1+e^x)(2+e^x)}[/Tex]

Answer:

[Tex]\int \frac{e^x}{(1+e^x)(2+e^x)} Let \ e^x = t e^x dx = dt\\ \Rightarrow \int \frac{e^x}{(1+e^x)(2+e^x)} dx = \int \frac{dt}{(t+1)(t+2)} \\= \int [\frac{1}{(t+1)} – \frac{1}{(t+2)} ] dt\\ = log|t+1|-log|t+2|+C\\ = log |\frac{t+1}{t+2}|+C\\ = log |\frac{1+e^x}{2+e^x}|+C[/Tex]

Question 14: [Tex]\frac{1}{(x^2 + 1)(x^2 + 4)}[/Tex]

Answer:

[Tex] \frac{1}{(x^2 + 1)(x^2 + 4)} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{x^2 + 4} \\ \Rightarrow 1 = (Ax + B)(x^2 + 4) + (Cx + D)(x^2 + 1) \\ \Rightarrow 1 = Ax^3 + 4Ax + Bx^2 + 4B + Cx^3 + Cx + Dx^2 + D \\Equating\ the\ coefficients\ of\ (x^3), (x^2), (x), and\ the\ constant\ term, we\ obtain\\ \begin{aligned} A + C &= 0 \\ B + D &= 0 \\ 4A + C &= 0 \\ 4B + D &= 1 \end{aligned} \\On\ solving\ these\ equations, \ we\ obtain \\ A = 0, \quad B = \frac{1}{3}, \quad C = 0, \quad \text{and} \quad D = -\frac{1}{3} \\From\ equation\ (1), \ we \ obtain \\\frac{1}{(x^2 + 1)(x^2 + 4)} = \frac{1}{3(x^2 + 1)} – \frac{1}{3(x^2 + 4)}\\ \\\int \frac{1}{(x^2 + 1)(x^2 + 4)} \, dx = \frac{1}{3} \int \frac{1}{x^2 + 1} \, dx – \frac{1}{3} \int \frac{1}{x^2 + 4} \, dx \\= \frac{1}{3} \tan^{-1} x – \frac{1}{3} \int \frac{1}{(x/2)^2 + 1} \, dx \\= \frac{1}{3} \tan^{-1} x – \frac{1}{3} \cdot \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C \\= \frac{1}{3} \tan^{-1} x – \frac{1}{6} \tan^{-1} \left( \frac{x}{2} \right) + C [/Tex]

Question 15: [Tex]\cos^3 x e^{\log \sin x}[/Tex]

Answer:

[Tex]\cos^3 x e^{\log \sin x} = \cos^3 x \times \sin x\\ Let\ \cos x = t \\ \implies-\sin x dx = dt\\ \implies \int cos^3xe^{logsin x}dx \\= \int cos^3xsin x dx\\ =-\int t^3 dt =-\frac{t^4}{4} + C\\ =-\frac{cos^4 x}{4} +C [/Tex]

Question 16: [Tex]e^{3log x}(x^4 + 1)^{-1}[/Tex]

Answer:

[Tex]e^{3log x}(x^4 + 1)^{-1}= e^{log x^3}(x^4 + 1)^{-1}=\frac{x^3}{(x^4+1)}\\ Let x^4+1=t \implies 4x^3 dx=dt\\ \implies \int e^{3log x}(x^4 +1)^{-1}dx=\int \frac{x^3}{(x^4+1)}dx\\ =\frac{1}{4}\int \frac{dt}{t}\\ =\frac{1}{4}log|t|+C\\ =\frac{1}{4}log|x^4+1|+C\\ =\frac{1}{4}log(x^4+1)+C [/Tex]

Question 17: [Tex]f'(ax+b)[f(ax+b)]^n[/Tex]

Answer:

[Tex]f'(ax+b)[f(ax+b)]^n\\ Let f(ax+b)=t \implies af'(ax+b)dx = dt\\ \implies \int f'(ax+b)[f(ax+b)]^n dx = \frac{1}{a}\int t^n dt \\= \frac{1}{a}\left[\frac{t^{n+1}}{n+1}\right]\\ = \frac{1}{a(n+1)}(f(ax+b))^{n+1}+C[/Tex]

Question 18: [Tex]\frac{1}{\sqrt{\sin^3 x \sin (x + \alpha)}}[/Tex]

Answer:

:[Tex] \frac{1}{\sqrt{\sin^3 x \sin (x + \alpha)}} = \frac{1}{\sqrt{\sin^3 x (\sin x \cos \alpha + \cos x \sin \alpha)}} \\= \frac{1}{\sqrt{\sin^4 x \cos \alpha + \sin^3 x \cos x \sin \alpha}} \\= \frac{1}{\sin^2 x \sqrt{\cos \alpha + \cot x \sin \alpha}} \\= \frac{\cosec^2 x}{\sqrt{\cos \alpha + \cot x \sin \alpha}} \\Let \ cos \alpha + cot x sin \alpha = t\\ \Rightarrow -\cosec^2 x sin \alpha \, dx = dt \\\therefore \int \frac{1}{\sqrt{sin^3 x sin (x + \alpha)}} \, dx = \int \frac{cosec^2 x \, dx}{\sqrt{cos \alpha + cot x sin \alpha}} \\= -\frac{1}{sin \alpha} \int \frac{dt}{\sqrt{t}} \\= -\frac{1}{sin \alpha} \left[ 2 \sqrt{t} \right] + C \\= -\frac{1}{sin \alpha} \left[ 2 \sqrt{cos \alpha + cot x sin \alpha} \right] + C \\= -\frac{2}{sin \alpha} \sqrt {cos \alpha +\frac{ cos x sin \alpha}{sin x}} + C \\= -\frac{2}{sin \alpha} \sqrt{\frac{sin x cos \alpha + cos x sin \alpha}{sin x}} + C \\= -\frac{2}{sin \alpha} \sqrt{\frac{sin (x + \alpha)}{sin x}} + C [/Tex]

Question 19: [Tex] \frac{\sin^{-1} \sqrt{x} – \cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} + \cos^{-1} \sqrt{x}} \, \, \in[0,1][/Tex]

Answer:

[Tex]I = \int \frac{\sin^{-1} \sqrt{x} – \cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} + \cos^{-1} \sqrt{x}} \, dx \\ \text{It is known that, } \sin^{-1} \sqrt{x} + \cos^{-1} \sqrt{x} = \frac{\pi}{2} \\ \Rightarrow I = \int \left( \frac{\frac{\pi}{2} – \cos^{-1} \sqrt{x} – \cos^{-1} \sqrt{x}}{\frac{\pi}{2}} \right) \, dx \\ = \frac{2\pi}{\pi} \int \left( \frac{\frac{\pi}{2} – 2 \cos^{-1} \sqrt{x}}{\pi} \right) \, dx \\ = \frac{2\pi}{\pi} \left( \frac{1}{2} \int \, dx – \frac{4}{\pi} \int \cos^{-1} \sqrt{x} \, dx \right) \\ = x – \frac{4}{\pi} \int \cos^{-1} \sqrt{x} \, dx \quad \ldots (1) \\ \text{Let } I_1 = \int \cos^{-1} \sqrt{x} \, dx \\ \text{Also, let } \sqrt{x} = t \Rightarrow dx = 2t \, dt \\ \Rightarrow I_1 = 2 \int \cos^{-1} t \cdot t \, dt \\ = 2 \left[ t^2 \cos^{-1} t – \int t^2 \, dt \right] \\ = t^2 \cos^{-1} t + \int \frac{t^2}{\sqrt{1 – t^2}} \, dt \\ = t^2 \cos^{-1} t – \int \frac{t^2}{\sqrt{1 – t^2}} \, dt \\ = t^2 \cos^{-1} t – \int t \, dt + \int t \sqrt{1 – t^2} \, dt \\ = t^2 \cos^{-1} t – \frac{t}{2} \sqrt{1 – t^2} + \frac{1}{2} \sin^{-1} t \\ = t^2 \cos^{-1} t – \frac{t}{2} \sqrt{1 – t^2} + \frac{1}{2} \sin^{-1} t \\From \ equation\ (1), we\ obtain \\ I = x – \frac{4}{\pi} \left[ t^2 \cos t – \frac{t}{2} \sqrt{1 – t^2} + \frac{1}{2} \sin^{-1} t \right] \\ = x – \frac{4}{\pi} \left[ x \cos^{-1} \sqrt{x} – \frac{\sqrt{x}}{2} \sqrt{1 – x} + \frac{1}{2} \sin^{-1} \sqrt{x} \right] \\ = x – \frac{4 x}{\pi} \left( \frac{-\sin^{-1} \sqrt{x}}{2} \right) – \frac{\sqrt{x} \cdot x^{1/2}}{2} + \sin^{-1} \sqrt{x} \\ = x – 2x + \frac{4x}{\pi} \sin^{-1} \sqrt{x} + \frac{2}{\pi} \sqrt{x – x^2} – \frac{2}{\pi} \sin^{-1} \sqrt{x} \\ = -x + \frac{2}{\pi} \left[ (2x – 1) \sin^{-1} \sqrt{x} \right] + \frac{2}{\pi} \sqrt{x – x^2} + C \\ = \frac{2(2x – 1)}{\pi} \sin^{-1} \sqrt{x} + \frac{2}{\pi} \sqrt{x – x^2} – x + C [/Tex]

Question 20: [Tex] \frac{\sqrt{1 – \sqrt{x}}}{\sqrt{1 + \sqrt{x}}} \, [/Tex]

Answer:

[Tex] \\I = \int \frac{\sqrt{1 – \sqrt{x}}}{\sqrt{1 + \sqrt{x}}} \, dx \\\text{Let } x = \cos^2 \theta \\\Rightarrow dx = -2 \sin \theta \cos \theta \, d\theta \\I = \int \frac{\sqrt{1 – \cos \theta}}{\sqrt{1 + \cos \theta}} (-2 \sin \theta \cos \theta) \, d\theta \\= -\int \frac{\sqrt{2 \sin^2 \frac{\theta}{2}}}{\sqrt{2 \cos^2 \frac{\theta}{2}}} \sin 2\theta \, d\theta \\= -\int \tan \frac{\theta}{2} \cdot 2 \sin \theta \cos \theta \, d\theta \\ = -2 \int \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \left( 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \right) \cos \theta \, d\theta\\ = -2 \int \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \left( 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \right) \cos \theta \, d\theta \\ = -4 \int \sin^2 \frac{\theta}{2} \cos \theta \, d\theta \\ = -4 \int \sin^2 \frac{\theta}{2} \left( 2 \cos^2 \frac{\theta}{2} – 1 \right) d\theta \\ = -4 \int \left( 2 \sin^2 \frac{\theta}{2} \cos^2 \frac{\theta}{2} – \sin^2 \frac{\theta}{2} \right) d\theta \\ = -8 \int \sin^2 \frac{\theta}{2} \cos^2 \frac{\theta}{2} \, d\theta + 4 \int \sin^2 \frac{\theta}{2} \, d\theta \\ = -2 \int \sin^2 \theta \, d\theta + 4 \int \sin^2 \frac{\theta}{2} \, d\theta \\ = -2 \int \left( \frac{1 – \cos 2\theta}{2} \right) d\theta + 4 \int \left( \frac{1 – \cos \theta}{2} \right) d\theta \\ = -2 \left[ \frac{\theta – \sin 2\theta}{4} \right] + 4 \left[ \frac{\theta – \sin \theta}{2} \right] + C \\ = -\frac{\theta – \sin 2\theta}{2} + 2\theta – 2 \sin \theta + C \\ = \theta + \frac{\sin 2\theta}{2} – 2 \sin \theta + C \\ = \theta + \frac{2 \sin \theta \cos \theta}{2} – 2 \sin \theta + C \\ = \theta + \sqrt{1 – \cos^2 \theta} \cdot \cos \theta – 2 \sqrt{1 – \cos^2 \theta} + C \\ = \cos^{-1} \sqrt{x} + \sqrt{1 – x} \cdot \sqrt{x} – 2 \sqrt{1 – x} + C \\ = -2 \sqrt{1 – x} + \cos^{-1} \sqrt{x} + \sqrt{x (1 – x)} + C \\ = -2 \sqrt{1 – x} + \cos^{-1} \sqrt{x} + \sqrt{x – x^2} + C[/Tex]

Question 21: [Tex]\frac{2 + \sin 2x}{1 + \cos 2x} e^x \,[/Tex]

Answer:

[Tex] I = \int \left( \frac{2 + \sin 2x}{1 + \cos 2x} \right) e^x \, dx\\ = \int \left( \frac{2 + 2 \sin x \cos x}{2 \cos^2 x} \right) e^x \,dx\\ = \int \left( \frac{1 + \sin x \cos x}{\cos^2 x} \right) e^x \, dx\\ = \int (\sec^2 x + \tan x) e^x \, dx\\ Let \ f(x) = \tan x \\\implies f'(x) = \sec^2 x \\ \therefore I = \int \left[ f(x) + f'(x) \right] e^x \, dx\\ = e^x f(x) + C\\ = e^x \tan x + C [/Tex]

Question 22: [Tex]\frac{x^2 + x + 1}{(x+1)^2 (x+2)}[/Tex]

Answer:

[Tex]Let \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A}{(x+1)} + \frac{B}{(x+1)^2} + \frac{C}{(x+2)} \quad \ldots (1) \\\Rightarrow x^2 + x + 1 = A(x+1)(x+2) + B(x+1)^2 + C(x^2 + 2x + 1) \\\Rightarrow x^2 + x + 1 = A(x^2 + 3x + 2) + B(x^2 + 2x + 1) + C(x^2 + 2x + 1) \Rightarrow x^2 + x + 1 = (A + B + C)x^2 + (3A + 2B + 2C)x + (2A + B + C) \\Equating \ the\ coefficients\ of\ ( x^2, x, ) and\ constant\ term, \ we\ obtain\\ A + C = 1\\ 3A + B + 2C = 1\\ 2A + 2B + C = 1\\ On\ solving\ these\ equations,\ we\ obtain\\ A = -2, \, B = 1, \, \text{and} \, C = 3\\ From \ equation (1),\ we \ obtain \\ \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{-2}{(x+1)} + \frac{3}{(x+2)} + \frac{1}{(x+1)^2} \\\int \frac{x^2 + x + 1}{(x+1)^2 (x+2)} \, dx = -2 \int \frac{1}{x+1} \, dx + 3 \int \frac{1}{x+2} \, dx + \int \frac{1}{(x+1)^2} \, dx \\= -2 \log |x+1| + 3 \log |x+2| – \frac{1}{x+1} + C [/Tex]

Question 23: [Tex]\tan^{-1} \left( \sqrt{\frac{1-x}{1+x}} \right) \, [/Tex]

Answer:

[Tex] I = \tan^{-1} \left( \sqrt{\frac{1-x}{1+x}} \right) \, dx\\ Let\ x = \cos \theta\\ \implies dx = -\sin \theta \, d\theta \\ I = \int \tan^{-1} \left( \sqrt{\frac{1 – \cos \theta}{1 + \cos \theta}} \right) (-\sin \theta \, d\theta)\\ = -\int \tan^{-1} \left( \sqrt{\frac{2 \sin^2 \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}} \right) \sin \theta \, d\theta\\ \\ = -\int \tan^{-1} \left( \tan \frac{\theta}{2} \right) \sin \theta \, d\theta \\ = -\frac{1}{2} \int \theta \sin \theta \, d\theta \\ = -\frac{1}{2} \left[ \theta (-\cos \theta) – \int (-\cos \theta) \, d\theta \right] \\ = -\frac{1}{2} \left[ -\theta \cos \theta + \sin \theta \right] \\ = \frac{1}{2} \left[ \theta \cos \theta – \sin \theta \right] \\ = \frac{1}{2} \left[ \cos^{-1} x \cdot x – \sqrt{1 – x^2} \right] + C \\ = \frac{1}{2} \left( x \cos^{-1} x – \sqrt{1 – x^2} \right) + C [/Tex]

Question 24: [Tex] \frac{\sqrt{x^2 + 1} \left[ \log(x^2 + 1) – 2 \log x \right]}{x^4} [/Tex]

Answer:

[Tex]\frac{\sqrt{x^2 + 1} \left[ \log(x^2 + 1) – 2 \log x \right]}{x^4} = \frac{\sqrt{x^2 + 1}}{x^4} \left[ \log(x^2 + 1) – \log x^2 \right]\\ = \frac{\sqrt{x^2 + 1}}{x^4} \log \left( \frac{x^2 + 1}{x^2} \right)\\ = \frac{\sqrt{x^2 + 1}}{x^4} \log \left( 1 + \frac{1}{x^2} \right)\\ = \frac{1}{x^3} \frac{\sqrt{x^2 + 1}}{\sqrt{x^2}} \log \left( 1 + \frac{1}{x^2} \right)\\ = \frac{1}{x^3} \sqrt{1 + \frac{1}{x^2}} \log \left( 1 + \frac{1}{x^2} \right)\\ \text{Let } 1 + \frac{1}{x^2} = t \\\Rightarrow \frac{-2}{x^3} dx = dt\\ \therefore I = \int \frac{1}{x^3} \sqrt{1 + \frac{1}{x^2}} \log \left( 1 + \frac{1}{x^2} \right) dx\\ = -\frac{1}{2} \int \sqrt{t} \log t \, dt\\ = -\frac{1}{2} \int t^{1/2} \cdot \log t \, dt\\ \text{Integrating \ by \ parts, we \ obtain}\\ I = -\frac{1}{2} \left[ \log t \cdot \int t^{1/2} dt – \left( \frac{d}{dt} \log t \right) \int t^{1/2} dt \right]\\ = -\frac{1}{2} \left[ \log t \cdot \frac{t^{3/2}}{3/2} – \int \frac{1}{t} \cdot \frac{t^{3/2}}{3/2} dt \right]\\ = -\frac{1}{2} \left[ \frac{2}{3} t^{3/2} \log t – \frac{2}{3} \int t^{1/2} dt \right]\\ = -\frac{1}{2} \left[ \frac{2}{3} t^{3/2} \log t – \frac{4}{9} t^{3/2} \right]\\ = \frac{1}{3} \left[ t^{3/2} \log t + \frac{2}{9} t^{3/2} \right]\\ = \frac{1}{3} t^{3/2} \left[ \log t – \frac{2}{3} \right]\\ = \frac{1}{3} \left( 1 + \frac{1}{x^2} \right)^{3/2} \left[ \log \left( 1 + \frac{1}{x^2} \right) – \frac{2}{3} \right] + C\\ [/Tex]

Question 25: [Tex] \int_{\frac{\pi}{2}}^{\pi} e^x \left( \frac{1 – \sin x}{1 – \cos x} \right) dx [/Tex]

Answer:

[Tex] I = \int_{\frac{\pi}{2}}^{\pi} e^x \left( \frac{1 – \sin x}{1 – \cos x} \right) dx\\ = \int_{\frac{\pi}{2}}^{\pi} e^x \left( \frac{1 – 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx\\ = \int_{\frac{\pi}{2}}^{\pi} e^x \left( \frac{\csc^2 \frac{x}{2} – \cot \frac{x}{2}}{2} \right) dx\\ \text{Let } f(x) = -\cot \frac{x}{2}\\ \Rightarrow f'(x) = -\left(-\frac{1}{2} \csc^2 \frac{x}{2}\right) = \frac{1}{2} \csc^2 \frac{x}{2}\\ \therefore I = \int_{\frac{\pi}{2}}^{\pi} e^x \left( f(x) + f'(x) \right) dx\\ = \left[ e^x \cdot f(x) \right]_{\frac{\pi}{2}}^{\pi}\\ = -\left[ e^x \cdot \cot \frac{x}{2} \right]_{\frac{\pi}{2}}^{\pi}\\ = -\left[ e^{\pi} \cot \frac{\pi}{2} – e^{\frac{\pi}{2}} \cot \frac{\pi}{4} \right]\\ = -\left[ e^{\pi} \cdot 0 – e^{\frac{\pi}{2}} \cdot 1 \right]\\ = e^{\frac{\pi}{2}} [/Tex]

Question 26: [Tex]\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos^{4} x + \sin^{4} x} dx[/Tex]

Answer:

[Tex]Let I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos^{4} x + \sin^{4} x} dx \\ \\ \implies I = \int_{0}^{\frac{\pi}{4}} \frac{\tan x \sec^{2} x}{1 + \tan^{4} x} dx \\ \\ \implies I = \int_{0}^{\frac{\pi}{4}} \frac{\tan x \sec^{2} x}{1 + \tan^{4} x} dx \\Let\ \tan^{2} x = t \implies 2 \tan x \sec^{2} x dx = dt \\When\ x = 0, t = 0 \ and\ when\ x = \frac{\pi}{4}, t = 1\\ \therefore I = \frac{1}{2} \int_{0}^{1} \frac{dt}{1 + t^{2}}\\ = \frac{1}{2} [\tan^{-1} t ]_{0}^{1}\\ = \frac{1}{2} [\tan^{-1} 1 – \tan^{-1} 0]\\ = \frac{1}{2} [\frac{\pi}{4}]\\ = \frac{\pi}{8}[/Tex]

Question 27: [Tex]\int_0^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} dx[/Tex]

Answer:

[Tex]Let I = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} dx\\ \Rightarrow I = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos^2 x + 4(1-\cos^2 x)} dx\\ \Rightarrow I = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos^2 x + 4 – 4 \cos^2 x} dx\\ \Rightarrow I = -\frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4-3 \cos^2 x-4}{4-3 \cos^2 x} dx\\ \Rightarrow I = -\frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4-3 \cos^2 x}{4-3 \cos^2 x} dx + \frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4}{4-3 \cos^2 x} dx\\ \Rightarrow I = -\frac{1}{3} \int_0^{\frac{\pi}{2}} 1 dx + \frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4 \sec^2 x}{4 \sec^2 x – 3} dx\\ \Rightarrow I = -\frac{1}{3} [x]_0^{\frac{\pi}{2}} + \frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4 (\sec^2 x)}{4 (1+\tan^2 x) – 3} dx\\ \Rightarrow I = -\frac{\pi}{6} + \frac{2}{3} \int_0^{\frac{\pi}{2}} \frac{2 \sec^2 x}{1 + 4 \tan^2 x} dx …(1)\\ Consider, \int_0^{\frac{\pi}{2}} \frac{2 \sec^2 x}{1 + 4 \tan^2 x} dx\\ Let 2 \tan x = t \Rightarrow 2 \sec^2 x dx = dt\\ When \ x = 0, \ t = 0 \ and \ when\ x = \frac{\pi}{2}, t = \infty\\ \Rightarrow \int_0^{\frac{\pi}{2}} \frac{2 \sec^2 x}{1 + 4 \tan^2 x} dx = \int_0^{\infty} \frac{dt}{1 + t^2}\\ = [\tan^{-1} t]_0^{\infty}\\ = [\tan^{-1} (\infty) – \tan^{-1} (0)]\\ = \frac{\pi}{2}\\ Therefore,\ from \ (1), we\ obtain \\ I = -\frac{\pi}{6} + \frac{2}{3} [\frac{\pi}{2}] = \frac{\pi}{3} – \frac{\pi}{6} = \frac{\pi}{6}\\[/Tex]

Question 28: [Tex]\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx[/Tex]

Answer:

[Tex] Let I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx\\ \Rightarrow I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{- (-\sin 2x)}} dx\\ \Rightarrow I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{- (1 – 1 – 2\sin x\cos x)}} dx\\ \Rightarrow I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{1 – (\sin^2 x + \cos^2 x – 2\sin x\cos x)}} dx\\ \Rightarrow I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x + \cos x) dx}{\sqrt{1 – (\sin x – \cos x)^2}}\\ Let (\sin x – \cos x) = t \\\Rightarrow (\sin x + \cos x) dx = dt\\ When x = \frac{\pi}{6}, t = (\frac{1 – \sqrt{3}}{2}) and when x = \frac{\pi}{3}, t = (\frac{\sqrt{3} – 1}{2})\\ I = \int_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{dt}{\sqrt{1-t^2}}\\ \Rightarrow I = [\sin^{-1} t]_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}}\\ As \sqrt{1 – (-\frac{1}{t})^2} = \sqrt{1 – \frac{1}{t^2}}, therefore,\\ \Rightarrow I = 2\int_{0}^{\frac{\sqrt{3} – 1}{2}} \frac{dt}{\sqrt{1-t^2}} \\ = [2\sin^{-1}t]_{0}^{\frac{\sqrt{3}-1}{2}} \\ = 2\sin^{-1} \left(\frac{\sqrt{3}-1}{2}\right)[/Tex]

Question 29: [Tex]\int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}[/Tex]

Answer:

[Tex]Let I = \int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}\\ I = \int_{0}^{1} \frac{1}{(\sqrt{1+x}-\sqrt{x}) }.\frac{(\sqrt{1+x}+\sqrt{x})} {(\sqrt{1+x}+\sqrt{x})} dx\\ = \int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1+x-x} dx \\ = \int_{0}^{1} \sqrt{1+x} dx + \int_{0}^{1} \sqrt{x} dx \\ = \left[\frac{2}{3}(1+x)^{\frac{3}{2}}\right]_{0}^{1} + \left[\frac{2}{3}(x)^{\frac{3}{2}}\right]_{0}^{1} \\ = \frac{2}{3}[(2)^{\frac{3}{2}}-1] + \frac{2}{3}[1] \\ = \frac{2}{3}(2)^{\frac{3}{2}} \\ = \frac{2.2\sqrt{2}}{3} \\ = \frac{4\sqrt{2}}{3}[/Tex]

Question 30: [Tex] \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx [/Tex]

Answer:

[Tex]Let I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx\\ Also, \ let \sin x – \cos x = t \Rightarrow (\cos x + \sin x) dx = dt\\ When\ x=0,\ t=-1\ and\ when\ x=\frac{\pi}{4}, t=0\\ \implies (\sin x – \cos x)^2 = t^2 \\ \implies \sin^2 x + \cos^2 x – 2 \sin x \cos x = t^2 \\ \implies 1 – \sin 2x = t^2 \\ \implies \sin 2x = 1 – t^2\\ \therefore I = \int_{-1}^{0} \frac{dt}{9 + 16 (1 – t^2)} = \int_{-1}^{0} \frac{dt}{25 – 16t^2}\\ = \int_{-1}^{0} \frac{dt}{(5)^2 – (4t)^2} = \frac{1}{4} \int_{-1}^{0} \frac{1}{5} \left[ \frac{1}{5} \log \left| \frac{5 + 4t}{5 – 4t} \right| \right]_{-1}^{0} \\ = \frac{1}{40} \left[ \log (1) – \log \left| \frac{1}{9} \right| \right] = \frac{1}{40} \log 9 [/Tex]

Question 31: [Tex]\int_{0}^{\frac{\pi}{2}} \sin 2x \tan^{-1} (\sin x) \, dx[/Tex]

Answer:

[Tex]\text{Let } I = \int_{0}^{\frac{\pi}{2}} \sin 2x \tan^{-1} (\sin x) \, dx = \int_{0}^{\frac{\pi}{2}} 2 \sin x \cos x \tan^{-1} (\sin x) \, dx \\ \text{Also, let } \sin x = t \implies \cos x \, dx = dt \\ \text{When } x = 0, \ t = 0 \text{ and when } x = \frac{\pi}{2}, t = 1 \\ \implies I = 2 \int_{0}^{1} t \tan^{-1} (t) \, dt \quad \cdots (1)\\ Consider\\ \int t \tan^{-1} t \, dt = \tan^{-1} t \cdot \int t \, dt – \int \left( \frac{d}{dt} (\tan^{-1} t) \cdot \int t \, dt \right) dt \\ = \tan^{-1} t \cdot \frac{t^2}{2} – \int \left( \frac{1}{1 + t^2} \cdot \frac{t^2}{2} \right) dt\\ = \tan^{-1} t \cdot \frac{t^2}{2} – \frac{1}{2} \int \frac{t^2 + 1 – 1}{1 + t^2} \, dt\\ = \frac{t^2 \tan^{-1} t}{2} – \frac{1}{2} \int \left( 1 – \frac{1}{1 + t^2} \right) dt\\ = \frac{t^2 \tan^{-1} t}{2} – \frac{1}{2} \left( \int 1 \, dt – \int \frac{1}{1 + t^2} \, dt \right)\\ = \frac{t^2 \tan^{-1} t}{2} – \frac{1}{2} \left( t – \tan^{-1} t \right)\\ = \frac{t^2 \tan^{-1} t}{2} – \frac{t}{2} + \frac{\tan^{-1} t}{2}\\ \implies \int_{0}^{1} t \tan^{-1} t \, dt = \left[ \frac{t^2 \tan^{-1} t}{2} – \frac{t}{2} + \frac{\tan^{-1} t}{2} \right]_{0}^{1}\\ = \left[ \frac{1^2 \tan^{-1} 1}{2} – \frac{1}{2} + \frac{\tan^{-1} 1}{2} \right] – \left[ \frac{0^2 \tan^{-1} 0}{2} – \frac{0}{2} + \frac{\tan^{-1} 0}{2} \right] \\ = \left[ \frac{\pi/4}{2} – \frac{1}{2} + \frac{\pi/4}{2} \right] – \left[ 0 \right]\\ = \frac{1}{2} \left[ \frac{\pi}{2} – 1 \right]\\ From \ equation\ (1), we\ obtain\\ I = 2 \left[ \frac{\pi}{4} – \frac{1}{2} \right] = \frac{\pi}{2} – 1 [/Tex]

Question 32: [Tex]\int_0^\pi \frac{x \tan x}{\sec x + \tan x} \, dx[/Tex]

Answer:

[Tex] Let I = \int_0^\pi \frac{x \tan x}{\sec x + \tan x} \, dx \\ I = \int_0^\pi \frac{(\pi – x) \tan (\pi – x)}{\sec (\pi – x) + \tan (\pi – x)} \, dx \\ \Rightarrow I = \int_0^\pi \frac{-(\pi – x) \tan x}{-(\sec x + \tan x)} \, dx \\ \Rightarrow I = \int_0^\pi \frac{(\pi – x) \tan x}{\sec x + \tan x} \, dx \\ Adding\ (1) \ and \ (2), we\ obtain \\ 2I = \int_0^\pi \frac{\pi \tan x}{\sec x + \tan x} \, dx \\ \Rightarrow 2I = \pi \int_0^\pi \frac{\sin x}{\cos x (1 + \sin x)} \, dx \\ \Rightarrow 2I = \pi \int_0^\pi \frac{\sin x + 1 – 1}{1 + \sin x} \, dx \\ \Rightarrow 2I = \pi \int_0^\pi 1 \, dx – \pi \int_0^\pi \frac{1}{1 + \sin x} \, dx \\ \Rightarrow 2I = \pi \left[ x \right]_0^\pi – \pi \int_0^\pi \frac{1 – \sin x}{\cos^2 x} \, dx \\ \Rightarrow 2I = \pi^2 – \pi \left[ \tan x – \sec x \right]_0^\pi \\ \Rightarrow 2I = \pi^2 – \pi \left[ \tan \pi – \sec \pi – \tan 0 + \sec 0 \right] \\ \Rightarrow 2I = \pi^2 – \pi \left[ 0 – (-1) – 0 + 1 \right] \\ \Rightarrow 2I = \pi^2 – 2\pi \\ \Rightarrow 2I = \pi (\pi – 2) \\ \Rightarrow I = \frac{\pi}{2} (\pi – 2) [/Tex]

Question 33:[Tex]\int_1^4 \left[ |x-1| + |x-2| + |x-3| \right] dx[/Tex]

Answer:

[Tex]Let ( I = \int_1^4 \left[ |x-1| + |x-2| + |x-3| \right] dx )\\ \Rightarrow I = \int_1^4 |x-1| \, dx + \int_1^4 |x-2| \, dx + \int_1^4 |x-3| \, dx\\ I = I_1 + I_2 + I_3 \quad \ldots (1)\\ where, ( I_1 = \int_1^4 |x-1| \, dx ), ( I_2 = \int_1^4 |x-2| \, dx ), and ( I_3 = \int_1^4 |x-3| \, dx )\\ I_1 = \int_1^4 (x-1) \, dx\\ (x-1) \geq 0 \text{ for } 1 \leq x \leq 4\\ \therefore I_1 = \int_1^4 (x-1) \, dx\\ \Rightarrow I_1 = \left[ \frac{x^2}{2} – x \right]_1^4 \\ \Rightarrow I_1 = \left[ \frac{16}{2} – 4 \right] – \left[ \frac{1}{2} – 1 \right] = 8 – 4 – \frac{1}{2} + 1 = \frac{9}{2} \quad \ldots (2)\\ I_2 = \int_1^4 |x-2| \, dx\\ x-2 \geq 0 \text{ for } 2 \leq x \leq 4 \text{ and } x-2 \leq 0 \text{ for } 1 \leq x \leq 2\\ \therefore I_2 = \int_2^4 (x-2) \, dx + \int_1^2 (2-x) \, dx\\ \Rightarrow I_2 = \left[ \frac{x^2}{2} – 2x \right]_2^4 + \left[ 2x – \frac{x^2}{2} \right]_1^2\\ \Rightarrow I_2 = \left[ \frac{16}{2} – 8 \right] – \left[ \frac{4}{2} – 2 \right] + \left[ 4 – 2 \right] – \left[ 2 – \frac{1}{2} \right]\\ \Rightarrow I_2 = \left[ 8 – 8 \right] + \left[ 8 – 2 \right] – \left[ 4 – \frac{1}{2} \right]\\ \Rightarrow I_2 = \frac{1}{2} + 2 = \frac{5}{2} \quad \ldots (3)\\ I_3 = \int_1^4 |x-3| \, dx\\ x-3 \geq 0 \text{ for } 3 \leq x \leq 4 \text{ and } x-3 \leq 0 \text{ for } 1 \leq x \leq 3\\ \therefore I_3 = \int_3^4 (x-3) \, dx + \int_1^3 (3-x) \, dx\\ \Rightarrow I_3 = \left[ \frac{x^2}{2} – 3x \right]_3^4 + \left[ 3x – \frac{x^2}{2} \right]_1^3\\ \Rightarrow I_3 = \left[ \frac{16}{2} – 12 \right] – \left[ \frac{9}{2} – 9 \right] + \left[ 9 – 3 \right] – \left[ 3 – \frac{1}{2} \right]\\ \Rightarrow I_3 = \left[ 8 – 12 \right] + \left[ 9 – \frac{9}{2} \right] = [6-4] + \left[ \frac{1}{2} \right] = \frac{5}{2} \quad \ldots (4)\\ From\ equations\ (1),\ (2),\ (3),\ and\ (4),\ we\ obtain\\ I = \frac{9}{2} + \frac{5}{2} + \frac{5}{2} = \frac{19}{2}\\ [/Tex]

Question 34:[Tex]\int_0^1 \frac{dx}{x^2(x+1)} = \frac{2}{3} + \log \frac{2}{3}[/Tex]

Answer:

[Tex]Let ( I = \int_0^1 \frac{dx}{x^2(x+1)} )\\ Also, let\\ \frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}\\ \Rightarrow 1 = A x (x+1) + B (x+1) + C x^2\\ \Rightarrow 1 = A x^2 + A x + B x + B + C x^2\\ Equating\ the\ coefficients\ of\ ( x^2 ), ( x ), and \ constant \ term, we\ obtain\\ A + C = 0\\ A + B = 0\\ B = 1\\ On solving these equations, we obtain\\ A = -1, \, C = 1, \, \text{and} \, B = 1\\ \therefore \quad \frac{1}{x^2(x+1)} = \frac{-1}{x} + \frac{1}{x^2} + \frac{1}{(x+1)} \\ \implies I = \int_{1}^{3} \left( \frac{-1}{x} + \frac{1}{x^2} + \frac{1}{(x+1)} \right) \, dx \\ = \left[ -\log x – \frac{1}{x} + \log (x+1) \right]_1^3 \\ = \left[ \log \left( \frac{x+1}{x} \right) – \frac{1}{x} \right]_1^3 \\ = \left[ \log \left( \frac{4}{3} \right) – \frac{1}{3} – \log \left( \frac{2}{1} \right) + 1 \right] \\ = \log 4 – \log 3 – \log 2 + \frac{2}{3} \\ = \log 2 – \log 3 + \frac{2}{3} \\ = \log \left( \frac{2}{3} \right) + \frac{2}{3}[/Tex]

Question 35:[Tex]\int_0^1 x e^x \, dx = 1[/Tex]

Answer:

:[Tex] Given: (\int_0^1 x e^x \, dx)\\ To\ Prove: (\int_0^1 x e^x \, dx = 1) \\ Let (I = \int_0^1 x e^x \, dx) \\ Using \ product \ rule \ we \ get \\ \int u \, dv = u v – \int v \, du \\ \Rightarrow \int_0^1 x e^x \, dx = x \left. e^x \right|_0^1 – \int_0^1 e^x \, dx On \ integrating \\ \Rightarrow \int_0^1 x e^x \, dx = \left[ x e^x \right]_0^1 – \left[ e^x \right]_0^1 Now\ by\ applying\ the\ limits\ we\ get\\ \Rightarrow \left[ x e^x \right]_0^1 – \left[ e^x \right]_0^1 = \left[ 1 \cdot e^1 – 0 \cdot e^0 \right] – \left[ e^1 – e^0 \right] \\\Rightarrow \left[ e – 0 \right] – \left[ e – 1 \right] = e – (e – 1) = 1 \\ Therefore, (\int_0^1 x e^x \, dx = 1) \\ \Rightarrow \int_0^1 x e^x \, dx = \left[ x e^x \right]_0^1 – \left[ e^x \right]_0^1 \\ \Rightarrow \int_0^1 x e^x \, dx = \left[1 \cdot e^1 – 0 \cdot e^0 \right] – \left[ e^1 – e^0 \right] \\ \Rightarrow \int_0^1 x e^x \, dx = e – 0 – e + 1 \\ \Rightarrow \int_0^1 x e^x \, dx = 1 \\ Therefore L.H.S = R.H.S\\ Hence\ Proved. [/Tex]

Question 36:[Tex]\int_{-1}^{1} x^{17} \cos^4 x \, dx = 0[/Tex]

Answer:

[Tex]Let ( I = \int_{-1}^{1} x^{17} \cos^4 x \, dx ) \\ Also, let f(x) = x^{17} \cos^4 x \\ \Rightarrow f(-x) = (-x)^{17} \cos^4 (-x) = -x^{17} \cos^4 x = -f(x) \\ Therefore, ( f(x) )\ is\ an\ odd\ function. \\ It\ is\ known\ that\ if\ f(x)\ is\ an\ odd\ function,\ then \\ \int_{-a}^{a} f(x) \, dx = 0 \\Hence I = \int_{-1}^{1} x^{17} \cos^4 x \, dx = 0 [/Tex]

Question 37:[Tex] \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx = \frac{2}{3} [/Tex]

Answer:

[Tex] Given: \\ \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx \\ To\ Prove: \\ \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx = \frac{2}{3} \\ Let \\ I = \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx \quad \ldots \quad (1) \\ The\ above\ equation \ can\ be\ written \ as: \\ I = \int_{0}^{\frac{\pi}{2}} \sin x \cdot \sin^2 x \, dx \\ = \int_{0}^{\frac{\pi}{2}} \sin x \cdot (1 – \cos^2 x) \, dx \\ Now\ by\ splitting\ the\ integrals: \\ I = \int_{0}^{\frac{\pi}{2}} \sin x \, dx – \int_{0}^{\frac{\pi}{2}} \sin x \cdot \cos^2 x \, dx \\ I = \left[-\cos x \right]_{0}^{\frac{\pi}{2}} – I_1 \quad \ldots \quad (2) \\ First, solve\ for (I_1): \\ I_1 = \int_{0}^{\frac{\pi}{2}} \sin x \cdot \cos^2 x \, dx \\ Let \cos x = t\\ \Rightarrow -\sin x \, dx = dt \\\Rightarrow \sin x \, dx = -dt) \\ When (x = 0) then\ (t = 1) and\ when\ \\(x = \frac{\pi}{2}) then (t = 0) \\ \Rightarrow I_1 = \int_{1}^{0} t^2 (-dt) \\ = -\int_{0}^{1} t^2 \, dt \\ On\ integrating\ we\ get \\ = -\left[ \frac{t^3}{3} \right]_{0}^{1} \\ Now\ by\ applying\ the\ limits\ we\ get \\ = -\left( \frac{1}{3} \right) \\ \Rightarrow I_1 = \frac{1}{3} \\ Substitute\ in\ equation\ (2) \\ \Rightarrow I = \left[ -\cos x \right]_{0}^{\frac{\pi}{2}} – \frac{1}{3} \\ \Rightarrow I = -\left( \cos \frac{\pi}{2} – \cos 0 \right) – \frac{1}{3} \\ \Rightarrow I = -\left( 0 – 1 \right) – \frac{1}{3} \\ \Rightarrow I = 1 – \frac{1}{3} \\ \Rightarrow I = \frac{2}{3} \\ L.H.S = R.H.S \\ Hence Proved. [/Tex]

Question 38:[Tex]\int_0^{\pi/4} 2 \tan^3 x \, dx = 1 – \log 2[/Tex]

Answer:

[Tex]\text{Let } I = \int_0^{\pi/4} 2 \tan^3 x \, dx\\ I = 2 \int_0^{\pi/4} \tan^2 x \tan x \, dx = 2 \int_0^{\pi/4} (\sec^2 x – 1) \tan x \, dx \\ = 2 \left[ \int_0^{\pi/4} \sec^2 x \tan x \, dx – \int_0^{\pi/4} \tan x \, dx \right] \\ = 2 \left[ \frac{\tan^2 x}{2} \right]_0^{\pi/4} + 2 \left[ \log \cos x \right]_0^{\pi/4} \\ = 1 + 2 \left[ \log \cos \frac{\pi}{4} – \log \cos 0 \right] \\ = 1 + 2 \left[ \log \frac{1}{\sqrt{2}} – \log 1 \right] \\ = 1 – \log 2 – \log 1 \\ = 1 – \log 2[/Tex]

Question 39:[Tex]\int_0^1 \sin^{-1} x \, dx = \frac{\pi}{2} – 1[/Tex]

Answer:

[Tex]\text{Let } I = \int_0^1 \sin^{-1} x \, dx\\ \implies I = \int_0^1 \sin^{-1} x \cdot 1 \, dx\\ Integrating\ by\ parts, we\ obtain\\ I=[\sin^{-1}x.x]_0^1 – \int_0^1 \frac{1}{\sqrt{1-x^2}} \cdot x dx\\ = [x\sin^{-1}x]_0^1 + \frac{1}{2}\int_0^1 \frac{(-2x)}{\sqrt{1-x^2}} dx \\Let 1-x^2 = t \\\implies -2x dx = dt\\ When\ x=0, t=1 and\ when\ x=1, t=0 I = [x\sin^{-1}x]_0^1 + \frac{1}{2}\int_1^0 \frac{dt}{\sqrt{t}}\\ = [x\sin^{-1}x]_0^1 + \frac{1}{2}[2\sqrt{t}]_1^0\\ = \sin^{-1}(1)+[-\sqrt{1}]\\ = \frac{\pi}{2}-1\\ Hence, the\ given\ result \ is\ proved.[/Tex]

Question 40: Evaluate [Tex]\int_0^1 e^{2-3x} \, dx[/Tex] as a limit of a sum.

Answer:

[Tex] Let \ I = \int_0^1 e^{2-3x} \, dx\\We \ know\ \int_a^b dx = (b-a)lim_{n\to\infty}\frac{1}{n}[f(a)+f(a+h)+…..+f(a+(n-1)h)]\\where, h = \frac{b-a}{n}\\Here, a=0, b=1 and f(x)=e^{2-3x}\\Hence, h =\frac{1-0}{n} = \frac{1}{n}\\Putting\ the\ values\ we\ get\\ \int_0^1 e^{2-3x} \, dx = (1-0) lim_{n\to\infty}\frac{1}{n}[f(0)+f(0+h)+…..+f(0+(n-1)h)]\\lim_{n \to \infty} \frac{1}{n} \left[ e^2 \left\{ 1 + e^{-3h} + e^{-6h} + e^{-9h} + \ldots e^{-3(n-1)h} \right\} \right] \\ = \lim_{n \to \infty} \frac{1}{n} \left[ e^2 \left( \frac{1 – \left( e^{-3h} \right)^n}{1 – \left( e^{-3h} \right)} \right) \right] \\ = \lim_{n \to \infty} \frac{1}{n} \left[ e^2 \left( \frac{1 – e^{-\frac{3h n}{n}}}{1 – e^{-3h}} \right) \right] \\ = \lim_{n \to \infty} \frac{1}{n} \left[ e^2 \left( \frac{1 – e^{-\frac{3}{n}}}{1 – e^{-3h}} \right) \right] \\ = \lim_{n \to \infty} \frac{1}{n} \left[ \frac{e^2 \left( 1 – e^{-3} \right)}{1 – e^{-\frac{3}{n}}} \right] \\ = e^2 \left( e^{-3} – 1 \right) \lim_{n \to \infty} \frac{1}{n} \left[ \frac{1}{\frac{3}{n} e^{-\frac{3}{n}}} \right] \\ = e^2 \left( e^{-3} – 1 \right) \lim_{n \to \infty} \left( \frac{1}{3} \right) \left[ \frac{-\frac{3}{n}}{e^{\frac{3}{n}} – 1} \right] \\ = -\frac{e^2 \left( e^{-3} – 1 \right)}{3} \lim_{n \to \infty} \left[ \frac{- \frac{3}{n}}{e^{\frac{3}{n}} – 1} \right] \\ = -\frac{e^2 \left( e^{-3} – 1 \right)}{3} (1) \\ = \frac{-e^{-1} + e^2}{3} \\ = \frac{1}{3} \left( e^2 – \frac{1}{e} \right) \left[ \lim_{n \to \infty} \frac{x}{e^x – 1} \right] [/Tex]

Question 41: [Tex]\int \frac{dx}{e^x + e^{-x}} \\(A) \tan^{-1}(e^x) + C \\(B) \tan^{-1}(e^{-x}) + C \\(C) \log(e^x – e^{-x}) + C \\(D) \log(e^x + e^{-x}) + C[/Tex]

Answer:

[Tex] \\ Given: \\ \int \frac{dx}{e^x + e^{-x}} \\ Let\ I = \int \frac{dx}{e^x + e^{-x}} \\ The\ above\ equation\ can\ be\ written\ as \\ I = \int \frac{dx}{e^{-x} (e^{2x} + 1)} \\ \\ = \int \frac{e^x dx}{e^{2x} + 1} \\ Put ( e^x = t \Rightarrow e^x dx = dt ) \\ \Rightarrow \int \frac{e^x dx}{e^{2x} + 1} = \int \frac{dt}{t^2 + 1} \\ \\ = \tan^{-1} t + C \\ = \tan^{-1} (e^x) + C \\ Hence, correct\ option\ is\ (A). [/Tex]

Question 42: [Tex]\int \frac{\cos 2x}{(\cos x + \sin x)^2} dx\\ \\ \text{(A) } \frac{-1}{\sin x + \cos x} + C \\ \text{(B) } \log|\sin x + \cos x| + C \\ \text{(C) } \log|\sin x – \cos x| + C \\ \text{(D) } \frac{1}{(\sin x + \cos x)^2} [/Tex]

Answer:

[Tex]\text{Let } I = \frac{\cos 2x}{(\cos x + \sin x)^2} \\ I = \int \frac{\cos^2 x – \sin^2 x}{(\cos x + \sin x)^2} \, dx \\ = \int \frac{(\cos x + \sin x)(\cos x – \sin x)}{(\cos x + \sin x)^2} \, dx \\ = \int \frac{\cos x – \sin x}{\cos x + \sin x} \, dx \\ \text{Let } \cos x + \sin x = t \implies (\cos x – \sin x) \, dx = dt \\ \therefore I = \int \frac{dt}{t} \\ = \log |t| + C \\ = \log |\cos x + \sin x| + C\\ Hence, \ the \ correct\ answer\ is\ B.[/Tex]

Question 43:[Tex]\left( \int_a^b f(x) dx = \int_a^b f(a + b – x)dx \right) If \ f(a+b-x)=f(x), then \int_a^b xf(x)dx\ is\ equal\ to (A) \frac{a+b}{2}\int_a^b f(b-x)dx (C) \frac{b-a}{2}\int_a^b f(x)dx (B) \frac{a+b}{2}\int_a^b f(b+x)dx (D) \frac{a+b}{2}\int_a^b f(x) dx[/Tex]

Answer:

[Tex]I = \int_a^b (a + b – x) f(a + b – x) dx\\ \Rightarrow I = \int_a^b (a + b – x)f(x) dx \Rightarrow I = (a + b) \int_a^b f(x)dx – 1\\ \Rightarrow I + I = (a + b) \int_a^b f(x)dx\\ \Rightarrow 2I = (a + b) \int_a^b f(x) dx\\ \Rightarrow I = \left( \frac{a + b}{2} \right) \int_a^b f(x) dx Hence, \\the\ correct\ answer\ is\ D. [/Tex]

Question 44: The value of [Tex]\int_0^1 tan^{-1}(\frac{2x-1}{1+x-x^2})dx is:[/Tex]

(A) 1

(B) 0

(C) -1

(D) [Tex]\frac{\pi}{4}[/Tex]

Answer:

[Tex]Let\ I = \int_0^1 tan^{-1}( \,\frac{2x-1}{1+x-x^2})dx\\ \implies I = \int_0^1 tan^{-1}( \,\frac{x-(1-x)}{1+x(1-x)})dx\\ \implies I = \int_0^1[tan^{-1} x – tan^{-1}(1-x)]dx ……(1)\\ \implies I = \int_0^1[tan^{-1} (1-x) – tan^{-1}(1-1+x)]dx \\ \implies I = \int_0^1[tan^{-1} (1-x) – tan^{-1}(x)]dx …….(2) \\ Add\ 1\ and\ 2, we\ get:\\ 2I = \int_0^1[tan^{-1}(x) + tan^{-1} (1-x) – tan^{-1} (1-x) – tan^{-1}(x)]dx\\ \implies 2I = 0\\ \implies I =0\\ Hence, the\ correct\ answer\ is\ B.[/Tex]

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