Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.7

In the article, we will solve Exercise 7.7 from Chapter 7, “Integrals” in the NCERT. Exercise 7.7 covers some special types of standard integrals which are based on integration by parts

Integral Formulae To Solve Exercise 7.7

[Tex]\int \sqrt{a^2-x^2}dx \space = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})+c[/Tex]…(i)

[Tex]\int \sqrt{x^2+a^2}dx \space = \frac{x}{2} \sqrt{x^2+a^2} + \frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}|+c[/Tex]…(iI)

[Tex]\int \sqrt{x^2-a^2}dx \space = \frac{x}{2} \sqrt{x^2-a^2} – \frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|+c[/Tex]…(iII)

Q.1: [Tex]\int \sqrt{1-4x^2}dx[/Tex]

Solution:

[Tex]\int \sqrt{4-x^2}dx \space = \int \sqrt{2^2-x^2}dx[/Tex]

by putting a = 2 in formula (1), we get

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

[Tex]=\frac{x}{2} \sqrt{4-x^2} + 2\sin^{-1}(\frac{x}{2})+c[/Tex]

Q.2: [Tex]\int \sqrt{1-4x^2}dx[/Tex]

Solution:

[Tex]\int \sqrt{1-4x^2}dx \space = \int \sqrt{1-(2x)^2}dx[/Tex]

by putting [Tex]2x = y, 2dx = dy, dx = \frac{dy}{2}[/Tex]

[Tex]=\frac{1}{2}\int \sqrt{1-y^2}dy[/Tex]

now by using formula (1), we get

[Tex]=\frac{1}{2}[\frac{y}{2} \sqrt{1-y^2} + \frac{1^2}{2}\sin^{-1}(\frac{y}{1})]+c[/Tex]

now we put y = 2x

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

[Tex]=\frac{x}{2} \sqrt{1-4x^2} + \frac{1}{4}\sin^{-1}(2x)+c[/Tex]

Q.3: [Tex]\int \sqrt{x^2+4x+6}dx[/Tex]

Solution:

[Tex]\int \sqrt{x^2+4x+6}dx \space = \int \sqrt{x^2+4x+2^2+2}dx[/Tex]

[Tex]\int \sqrt{(x+2)^2+(\sqrt2)^2}dx[/Tex]

let (x+2) = y and dx = dy

[Tex]\int \sqrt{y^2+(\sqrt2)^2}dy[/Tex]

using formula (2) , we get

[Tex] = \frac{y}{2} \sqrt{y^2+(\sqrt2)^2} + \frac{ (\sqrt2)^2}{2}\ln|y+\sqrt{y^2+(\sqrt2)^2}|+c[/Tex]

now putting y = (x+2)

[Tex] = \frac{(x+2)}{2} \sqrt{(x+2)^2+(\sqrt2)^2} + \ln|x+2+\sqrt{(x+2)^2+(\sqrt2)^2}|+c[/Tex]

[Tex] = \frac{(x+2)}{2} \sqrt{x^2+4x+6} + \ln|x+2+\sqrt{x^2+4x+6}|+c[/Tex]

Q.4: [Tex]\int \sqrt{x^2+4x+1}dx [/Tex]

Solution:

[Tex]\int \sqrt{x^2+4x+1}dx \space = \int \sqrt{x^2+4x+4-4+1}dx[/Tex]

[Tex]=\int \sqrt{(x+2)^2-(\sqrt3)^2}dx[/Tex]

substitute (x+2) = y and dx = dy

[Tex]= \int \sqrt{y^2-(\sqrt3)^2}dy \space = \frac{y}{2} \sqrt{y^2-(\sqrt3)^2} – \frac{(\sqrt3)^2}{2}\ln|y+\sqrt{y^2-(\sqrt3)^2}|+c[/Tex]

[Tex] = \frac{(x+2)}{2} \sqrt{(x+2)^2-(\sqrt3)^2} – \frac{3}{2}\ln|x+2+\sqrt{(x+2)^2-(\sqrt3)^2}|+c[/Tex]

[Tex] = \frac{(x+2)}{2} \sqrt{x^2+4x+1} – \frac{3}{2}\ln|x+2+\sqrt{x^2+4x+1}|+c[/Tex]

Q.5: [Tex]\int \sqrt{1-4x-x^2}dx [/Tex]

Solution:

[Tex]\int \sqrt{1-4x-x^2}dx \space = \int \sqrt{-(x^2+4x-1)}dx[/Tex]

[Tex] = \int \sqrt{-(x^2+4x+4-4-1)}dx [/Tex]

[Tex] = \int \sqrt{-((x+2)^2-(\sqrt5)^2)}dx \space = \int \sqrt{(\sqrt5)^2-(x+2)^2}dx [/Tex]

now putting (x+2) = y and dx = dy

[Tex]\int \sqrt{(\sqrt5)^2-y^2}dy[/Tex]

using formula (1)

[Tex] = \frac{y}{2} \sqrt{(\sqrt5)^2-y^2} + \frac{(\sqrt5)^2}{2}\sin^{-1}(\frac{y}{\sqrt5})+c[/Tex]

now putting (x+2) =y

[Tex] = \frac{(x+2)}{2} \sqrt{(\sqrt5)^2-(x+2)^2} + \frac{5}{2}\sin^{-1}(\frac{x+2}{\sqrt5})+c[/Tex]

[Tex] = \frac{(x+2)}{2} \sqrt{1-4x-x^2} + \frac{5}{2}\sin^{-1}(\frac{x+2}{\sqrt5})+c[/Tex]

Q.6: [Tex]\int \sqrt{x^2+4x-5}dx [/Tex]

Solution:

[Tex]\int \sqrt{x^2+4x-5}dx \space = \int \sqrt{x^2+4x+4-4-5}dx[/Tex]

[Tex]\int \sqrt{(x+2)^2-3^2}dx [/Tex]

now substitute (x+2) = y and dx = dy

[Tex]\int \sqrt{y^2-3^2}dy \space = \frac{y}{2} \sqrt{y^2-3^2} – \frac{3^2}{2}\ln|y+\sqrt{y^2-3^2}|+c[/Tex]

now put (x+2) = y

[Tex]= \frac{(x+2)}{2} \sqrt{(x+2)^2-3^2} – \frac{9}{2}\ln|x+2+\sqrt{(x+2)^2-3^2}|+c[/Tex]

[Tex]= \frac{(x+2)}{2} \sqrt{x^2+4x-5} – \frac{9}{2}\ln|x+2+\sqrt{x^2+4x-5}|+c[/Tex]

Q.7: [Tex]\int \sqrt{1+3x-x^2}dx [/Tex]

Solution:

[Tex]\int \sqrt{1+3x-x^2}dx \space = \int \sqrt{-(x^2-3x-1)}dx [/Tex]

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

[Tex]= \int \sqrt{-((x-\frac{3}{2})^2-(\frac{\sqrt13}{2})^2)}dx [/Tex]

[Tex]= \int \sqrt{(\frac{\sqrt13}{2})^2-(x-\frac{3}{2})^2}dx [/Tex]

now substitute , [Tex](x-\frac{3}{2}) = y[/Tex] and dx = dy

[Tex]= \int \sqrt{(\frac{\sqrt13}{2})^2-y^2}dy [/Tex]

using formula (1), we get

[Tex]= \frac{y}{2} \sqrt{(\frac{\sqrt13}{2})^2-y^2} + \frac{(\frac{\sqrt13}{2})^2}{2}\sin^{-1}(\frac{y}{\frac{\sqrt13}{2}})+c[/Tex]

[Tex]= \frac{(x-\frac{3}{2})}{2} \sqrt{(\frac{\sqrt13}{2})^2-(x-\frac{3}{2})^2} + \frac{13}{8}\sin^{-1}(\frac{(x-\frac{3}{2})}{\frac{\sqrt13}{2}})+c[/Tex]

[Tex]= \frac{(2x-3)}{4} \sqrt{1+3x-x^2} + \frac{13}{8}\sin^{-1}(\frac{2x-3}{\sqrt13})+c[/Tex]

Q.8: [Tex]\int \sqrt{x^2+3x}dx [/Tex]

Solution:

[Tex]= \int \sqrt{x^2+3x+(\frac{3}{2})^2 – (\frac{3}{2})^2}dx [/Tex]

[Tex]= \int \sqrt{(x+\frac{3}{2})^2 – (\frac{3}{2})^2}dx [/Tex]

now substitute [Tex](x+\frac{3}{2}) = y [/Tex]and dx = dy

[Tex]= \int \sqrt{y^2 – (\frac{3}{2})^2}dy [/Tex]

using formula (3) , we get

[Tex] = \frac{y}{2} \sqrt{y^2-(\frac{3}{2})^2} – \frac{(\frac{3}{2})^2}{2}\ln|y+\sqrt{y^2-(\frac{3}{2})^2}|+c[/Tex]

[Tex] = \frac{(x+\frac{3}{2})}{2} \sqrt{(x+\frac{3}{2})^2-(\frac{3}{2})^2} – \frac{9}{2}\ln|x+\frac{3}{2}+\sqrt{(x+\frac{3}{2})^2-(\frac{3}{2})^2}|+c[/Tex]

[Tex] = \frac{(2x+3)}{4} \sqrt{x^2+3x} – \frac{9}{8}\ln|x+\frac{3}{2}+\sqrt{x^2+3x}|+c[/Tex]

Q.9: [Tex]\int \sqrt{1+(\frac{x^2}{9})}dx [/Tex]

Solution:

[Tex]\int \sqrt{1+(\frac{x^2}{9})}dx = \frac{1}{3}\int \sqrt{x^2+3^2)}dx [/Tex]

by using formula (2)

[Tex]\frac{1}{3}\int \sqrt{x^2+3^2}dx \space = \frac{1}{3}[\frac{x}{2} \sqrt{x^2+3^2} + \frac{3^2}{2}\ln|x+\sqrt{x^2+3^2}|]+c[/Tex]

[Tex] = \frac{x}{6} \sqrt{x^2+9} + \frac{3}{2}\ln|x+\sqrt{x^2+9}|+c[/Tex]

Q.10: [Tex]\int \sqrt{1+x^2}dx[/Tex]

(A) x/2√(1 + x²) + 1/2log|x + √(1 + x²) + C|

(B) 2/3(1 + x²)^3/2 + C

(C) 2/3x√(1 + x²)^3/2 + C

(D) x²/2√(1 + x²) + 1/2.x².log|x + √(1 + x²) + C|

Solution:

Option (A) is Correct

Using formula (2) we get

[Tex]\int \sqrt{x^2+1^2}dx \space = \frac{x}{2} \sqrt{x^2+1^2} + \frac{1^2}{2}\ln|x+\sqrt{x^2+1^2}|+c[/Tex]

[Tex] = \frac{x}{2} \sqrt{x^2+1} + \frac{1}{2}\ln|x+\sqrt{x^2+1}|+c[/Tex]

Q.11: [Tex]\int \sqrt{x^2-8x+7}dx [/Tex]

(A) 1/2(x – 4)√(x² – 8x + 7) + 9.log|x – 4 + √(x² -8x + 7)| + C

(B) 1/2(x – 4)√(x² – 8x + 7) + 9.log|x + 4 + √(x² -8x + 7)| + C

(C) 1/2(x – 4)√(x² – 8x + 7) – 3√(2).log|x – 4 + √(x² -8x + 7)| + C

(D) 1/2(x – 4)√(x² – 8x + 7) – 9/2.log|x – 4 + √(x² -8x + 7)| + C

Solution:

Option (D) is Correct

[Tex]\int \sqrt{x^2-8x+7}dx \space = \int \sqrt{x^2-8x+4^2-4^2+7}dx [/Tex]

[Tex] = \int \sqrt{(x-4)^2-3^2}dx [/Tex]

let (x-4) = y and dx = dy

[Tex]\int\sqrt{y^2-3^2}dy[/Tex]

using formula (3)

[Tex]\int \sqrt{y^2-3^2}dy \space = \frac{y}{2} \sqrt{y^2-3^2} – \frac{3^2}{2}\ln|y+\sqrt{y^2-3^2}|+c[/Tex]

Now put y = (x-4) and after simplifying

[Tex] = \frac{(x-4)}{2} \sqrt{x^2-8x+7} – \frac{9}{2}\ln|x-4+\sqrt{x^2-8x+7}|+c[/Tex]