Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.3

In this article, we will discuss all the solutions for Exercise 3 of Chapter 7 of Class 12 NCERT.

1. sin²(2x + 5)

Solution:

Let u = 2x + 5

Then, du = 2dx

1/2 ∫sin²(u) du

Using the identity sin²(u) = (1 – cos(2u))/2

= 1/2 ∫(1 – cos(2u))/2 du

= 1/4 ∫(1 – cos(2u)) du

= 1/4 (u – 1/2 sin(2u)) + C

= 1/4 (2x + 5 – 1/2 sin(4x + 10)) + C

2. sin(3x)cos(4x)

Solution:

Using the identity sin(a)cos(b) = 1/2(sin(a+b) + sin(a-b))

= 1/2 ∫(sin(7x) + sin(-x)) dx

= -1/14 cos(7x) + 1/2 sin(-x) + C

= -1/14 cos(7x) – 1/2 sin(x) + C

3. cos(2x)cos(4x)cos(6x)

Solution:

Using the product-to-sum formula: cos(a)cos(b)cos(c) = 1/4(cos(a+b+c) + cos(a+b-c) + cos(a-b+c) + cos(-a+b+c))

= 1/4 ∫(cos(12x) + cos(-2x) + cos(2x) + cos(10x)) dx

= 1/4 (1/12 sin(12x) + 1/2 sin(-2x) + 1/2 sin(2x) + 1/10 sin(10x)) + C

= 1/48 sin(12x) – 1/8 sin(2x) + 1/8 sin(2x) + 1/40 sin(10x) + C

= 1/48 sin(12x) + 1/40 sin(10x) + C

4. sin³(2x + 1)

Solution:

Let u = 2x + 1

Then, du = 2dx

1/2 ∫sin³(u) du

Using the reduction formula: ∫sinⁿ(u) du = -1/n sin^(n-1)(u)cos(u) + (n-1)/n ∫sin^(n-2)(u) du

= -1/3 sin²(u)cos(u) + 2/3 ∫sin(u) du

= -1/3 (1 – cos²(u))cos(u) + 2/3 (-cos(u)) + C

= -1/3 (cos(u) – cos³(u)) – 2/3 cos(u) + C

= -1/3 cos(u) + 1/3 cos³(u) – 2/3 cos(u) + C

= -cos(u) – 1/3 cos³(u) + C

= -cos(2x + 1) – 1/3 cos³(2x + 1) + C

5. sin³(x)cos³(x)

Solution:

Using the identity sin³(x) = (3sin(x) – sin(3x))/4 and cos³(x) = (cos(3x) + 3cos(x))/4

= 1/16 ∫(3sin(x) – sin(3x))(cos(3x) + 3cos(x)) dx

= 1/16 ∫(3sin(x)cos(3x) + 9cos(x)sin(x) – sin(3x)cos(3x) – 3sin(3x)cos(x)) dx

= 1/16 (3∫sin(x)cos(3x) dx + 9∫cos(x)sin(x) dx – ∫sin(3x)cos(3x) dx – 3∫sin(3x)cos(x) dx)

= 1/16 (-3/2 cos(3x) + 9/2 sin(2x) – 1/2 sin(6x) – 3/2 cos(4x)) + C

= -3/32 cos(3x) + 9/32 sin(2x) – 1/32 sin(6x) – 3/32 cos(4x) + C

6. sin(x)sin(2x)sin(3x)

Solution:

Using the identity sin(a)sin(b) = 1/2(cos(a-b) – cos(a+b))

= 1/2 ∫(cos(x-2x) – cos(x+2x))(1/2(cos(x-3x) – cos(x+3x))) dx

= 1/4 ∫(cos(-x) – cos(3x) – cos(x) + cos(5x)) dx

= 1/4 (-sin(x) – 1/3 sin(3x) – sin(x) + 1/5 sin(5x)) + C

= -1/2 sin(x) – 1/12 sin(3x) + 1/5 sin(5x) + C

7. sin(4x)sin(8x)

Solution:

Using the identity sin(a)sin(b) = 1/2(cos(a-b) – cos(a+b))

= 1/2 ∫(cos(4x-8x) – cos(4x+8x)) dx

= 1/2 ∫(cos(-4x) – cos(12x)) dx

= 1/2 (-1/4 sin(4x) – 1/12 sin(12x)) + C

= -1/8 sin(4x) – 1/24 sin(12x) + C

8. (1 – cos(x) /1+cos(x))

Solution:

1. We’ll use the trigonometric identity [Tex]\cos^2(x) + \sin^2(x) = 1[/Tex] to simplify the expression:

[Tex]frac{1 – \cos(x)}{1 + \cos(x)} = \frac{1 – \cos(x)}{1 + \cos(x)} \cdot \frac{1 – \cos(x)}{1 – \cos(x)} = \frac{(1 – \cos(x))^2}{1 – \cos^2(x)} = \frac{(1 – \cos(x))^2}{\sin^2(x)}[/Tex]

2. Now, we’ll rewrite the integral in terms of this simplified expression:

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

3. Next, we’ll expand the numerator [Tex]((1 – \cos(x))^2[/Tex]:

[Tex](1 – \cos(x))^2 = 1 – 2\cos(x) + \cos^2(x)[/Tex]

4. Now, our integral becomes:

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

5. We can further simplify this by breaking it into three separate integrals:

[Tex]\int \frac{1}{\sin^2(x)} \, dx – 2\int \frac{\cos(x)}{\sin^2(x)} \, dx + \int \frac{\cos^2(x)}{\sin^2(x)} \, dx[/Tex]

6. Now, we’ll integrate each term:

[Tex]- \int \frac{1}{\sin^2(x)} \, dx[/Tex] can be simplified using the trigonometric identity [Tex]\csc^2(x) = \frac{1}{\sin^2(x)}[/Tex]:

[Tex]= \int \csc^2(x) \, dx = -\cot(x) + C_1[/Tex]

– [Tex]\int \frac{\cos(x)}{\sin^2(x)} \, dx[/Tex] can be simplified using substitution:

Let [Tex]u = \sin(x)[/Tex], then [Tex]du = \cos(x) \, dx[/Tex].

The integral becomes:

[Tex]= -\int \frac{1}{u^2} \, du = -(-\cot(x)) + C_2 = \cot(x) + C_2[/Tex]

[Tex]- \int \frac{\cos^2(x)}{\sin^2(x)} \, dx[/Tex] can be simplified using the Pythagorean identity [Tex]\cos^2(x) = 1 – \sin^2(x)[/Tex]:

[Tex]= \int \frac{1 – \sin^2(x)}{\sin^2(x)} \, dx = \int \left(\frac{1}{\sin^2(x)} – 1\right) \, dx[/Tex]

[Tex]= -\cot(x) – x + C_3[/Tex]

7. Putting it all together, the final result is:

[Tex]-\cot(x) – \cot(x) – x + C = -2\cot(x) – x + C[/Tex]

So, [Tex]\int \frac{1 – \cos(x)}{1 + \cos(x)} \, dx = -2\cot(x) – x + C,[/Tex] where C is the constant of integration.

9. (cos(x)/(1 +(cos(x)))

Solution:

1. We’ll start by multiplying the numerator and denominator by [Tex](1 – \cos(x))[/Tex] to get rid of the denominator:

[Tex]\int \frac{\cos(x)}{1 + \cos(x)} \, dx = \int \frac{\cos(x)(1 – \cos(x))}{(1 + \cos(x))(1 – \cos(x))} \, dx[/Tex]

[Tex]= \int \frac{\cos(x) – \cos^2(x)}{1 – \cos^2(x)} \, dx[/Tex]

2. Using the Pythagorean identity [Tex]\sin^2(x) + \cos^2(x) = 1[/Tex], we can simplify the denominator:

[Tex]1 – \cos^2(x) = \sin^2(x)[/Tex]

[Tex]\int \frac{\cos(x) – \cos^2(x)}{\sin^2(x)} \, dx[/Tex]

3. We can further simplify this expression by breaking it into two separate integrals:

[Tex]\int \frac{\cos(x)}{\sin^2(x)} \, dx – \int \frac{\cos^2(x)}{\sin^2(x)} \, dx[/Tex]

4. For the first integral, we use the substitution [Tex]u = \sin(x), du = \cos(x) \, dx[/Tex]:

[Tex]\int \frac{\cos(x)}{\sin^2(x)} \, dx = \int \frac{1}{u^2} \, du = -\cot(x) + C_1[/Tex]

5. For the second integral, we use the Pythagorean identity [Tex]\cos^2(x) = 1 – \sin^2(x)[/Tex]:

[Tex]\int \frac{\cos^2(x)}{\sin^2(x)} \, dx = \int \frac{1 – \sin^2(x)}{\sin^2(x)} \, dx = \int \left(\frac{1}{\sin^2(x)} – 1\right) \, dx[/Tex]

[Tex]= -\cot(x) – x + C_2[/Tex]

[Tex]-\cot(x) + (-\cot(x) – x) + C = -2\cot(x) – x + C[/Tex]

[Tex]\int \frac{\cos(x)}{1 + \cos(x)} \, dx = -2\cot(x) – x + C[/Tex]

10. sin⁴(x)

Solution:

Using the identity sin²(x) = (1 – cos(2x))/2

= ∫(1 – cos(2x))²/4 dx

= 1/4 ∫(1 – 2cos(2x) + cos²(2x)) dx

= 1/4 ∫(1 – 2cos(2x) + (1 + cos(4x))/2) dx

= 1/4 ∫(3/2 – 2cos(2x) + 1/2cos(4x)) dx

= 1/4 (3/2x – sin(2x) + 1/8 sin(4x)) + C

= 3/8x – 1/4 sin(2x) + 1/32 sin(4x) + C

11. cos⁴(2x)

Solution:

Using the identity cos²(x) = (1 + cos(2x))/2

= ∫(1 + cos(4x))²/4 dx

= 1/4 ∫(1 + 2cos(4x) + cos²(4x)) dx

= 1/4 ∫(1 + 2cos(4x) + (1 + cos(8x))/2) dx

= 1/4 ∫(3/2 + 2cos(4x) + 1/2cos(8x)) dx

= 1/4 (3/2x + 1/2sin(4x) + 1/16 sin(8x)) + C

= 3/8x + 1/8 sin(4x) + 1/64 sin(8x) + C

12. (sin²(x)/(1+cos(x))

Solution:

1. We’ll start by using the Pythagorean identity [Tex]\sin^2(x) = 1 – \cos^2(x)[/Tex] to rewrite the integrand:

[Tex]\frac{\sin^2(x)}{1 + \cos(x)} = \frac{1 – \cos^2(x)}{1 + \cos(x)}[/Tex]

2. Next, we’ll rewrite the integral in terms of this simplified expression:

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

3. Now, we’ll simplify the expression further by dividing the numerator by [Tex]1 + \cos(x)[/Tex]:

[Tex]= \int \frac{1}{1 + \cos(x)} \, dx – \int \frac{\cos^2(x)}{1 + \cos(x)} \, dx[/Tex]

4. We can now integrate each term separately:

For the first integral, we use the substitution [Tex]u = \tan\frac{x}{2}[/Tex]:

[Tex]= \int \frac{1}{1 + \cos(x)} \, dx[/Tex]

Using the half-angle formula [Tex]\cos(x) = \frac{1 – \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}[/Tex], the integral becomes:

[Tex]= \int \frac{1}{2\cos^2\left(\frac{x}{2}\right)} \, dx [/Tex]

[Tex]= \int \sec^2\left(\frac{x}{2}\right) \, dx[/Tex]

[Tex]= 2\tan\left(\frac{x}{2}\right) + C_1[/Tex]

For the second integral, we’ll use a substitution [Tex]u = \sin(x), du = \cos(x) \, dx:[/Tex]

[Tex]= \int \frac{1 – u^2}{1 + u} \, du[/Tex]

[Tex]= \int \frac{1}{1 + u} \, du – \int u \, du[/Tex]

[Tex]= \ln|1 + u| – \frac{u^2}{2} + C_2[/Tex]

[Tex]2\tan\left(\frac{x}{2}\right) + \ln|1 + \sin(x)| – \frac{\sin^2(x)}{2} + C[/Tex]

[Tex]\int \frac{\sin^2(x)}{1 + \cos(x)} \, dx = 2\tan\left(\frac{x}{2}\right) + \ln|1 + \sin(x)| – \frac{\sin^2(x)}{2} + C[/Tex]

13. (cos(2x)-cos(2α)/cos(x)cos(α))

Solution:

1. We’ll start by using the double angle formula for cosine: [Tex]\cos(2\theta) = \cos^2(\theta) – \sin^2(\theta)[/Tex].

[Tex]\cos(2x) – \cos(2\alpha) = (\cos^2(x) – \sin^2(x)) – (\cos^2(\alpha) – \sin^2(\alpha))[/Tex]

[Tex]= \cos^2(x) – \cos^2(\alpha) – (\sin^2(x) – \sin^2(\alpha))[/Tex]

2. Next, we’ll use the identity [Tex]\sin^2(\theta) = 1 – \cos^2(\theta)[/Tex] to rewrite the expression:

[Tex]\cos(2x) – \cos(2\alpha) = \cos^2(x) – \cos^2(\alpha) – (1 – \cos^2(x) – (1 – \cos^2(\alpha)))[/Tex]

[Tex]= \cos^2(x) – \cos^2(\alpha) – 1 + \cos^2(x) + 1 – \cos^2(\alpha)[/Tex]

[Tex]= 2\cos^2(x) – 2\cos^2(\alpha)[/Tex]

3. Now, we’ll rewrite the integral in terms of this simplified expression:

[Tex]\int \frac{\cos(2x) – \cos(2\alpha)}{\cos(x) – \cos(\alpha)} \, dx = \int \frac{2\cos^2(x) – 2\cos^2(\alpha)}{\cos(x) – \cos(\alpha)} \, dx[/Tex]

4. We can further simplify the expression by factoring out a 2 from the numerator:

[Tex]= 2\int \frac{\cos^2(x) – \cos^2(\alpha)}{\cos(x) – \cos(\alpha)} \, dx[/Tex]

5. Now, we’ll use the difference of squares identity: [Tex]\cos^2(x) – \cos^2(\alpha) = (\cos(x) + \cos(\alpha))(\cos(x) – \cos(\alpha))[/Tex]

[Tex]= 2\int \frac{(\cos(x) + \cos(\alpha))(\cos(x) – \cos(\alpha))}{\cos(x) – \cos(\alpha)} \, dx[/Tex]

6. We can cancel out the common factor of in the numerator and denominator:

[Tex]= 2\int \cos(x) + \cos(\alpha) \, dx[/Tex]

7. Now, we can integrate each term separately:

[Tex]= 2\left(\int \cos(x) \, dx + \int \cos(\alpha) \, dx\right)[/Tex]

[Tex]= 2(\sin(x) + \cos(\alpha)x) + C[/Tex]

[Tex]\int \frac{\cos(2x) – \cos(2\alpha)}{\cos(x) – \cos(\alpha)} \, dx = 2(\sin(x) + \cos(\alpha)x) + C[/Tex]

14. (cos(x)-sin(x)/(1+sin(2x)))

Solution:

Using substitution method, let’s set u = sin(x), then du = cos(x) dx. The integral becomes:

∫(cos(x) – sin(x))/(1 + sin(2x)) dx = ∫(1 – u)/(1 + 2u√(1 – u^2)) du

Now, we can use partial fractions to simplify the integrand:

(1 – u)/(1 + 2u√(1 – u^2)) = A/(1 + u) + B√(1 – u^2)/(1 + u)

Multiplying both sides by the denominator, we get:

1 – u = A(1 + u) + B√(1 – u^2)

Now, substitute u = 0 to solve for A:

1 = A(1) => A = 1

Next, substitute u = 1 to solve for B:

0 = 1 + B√(0) => B = -1

Therefore, the integrand can be rewritten as:

∫(cos(x) – sin(x))/(1 + sin(2x)) dx = ∫(1 – u)/(1 + 2u√(1 – u^2)) du

= ∫(1/(1 + u) – √(1 – u^2)/(1 + u)) du

= ∫(1/(1 + u) du – ∫√(1 – u^2)/(1 + u) du

Now, we can solve these integrals separately:

1. ∫(1/(1 + u)) du:

This is a simple integral and equals ln|1 + u| + C.

2. ∫√(1 – u^2)/(1 + u) du:

Let’s make a substitution v = 1 – u^2, then dv = -2u du. The integral becomes:

-1/2 ∫√v dv = -1/2 * 2/3 v^(3/2) + C

= -1/3 (1 – u^2)^(3/2) + C

= ln|1 + sin(x)| – 1/3 (1 – sin^2(x))^(3/2) + C

15. tan³(2x)sec(2x)

Solution:

Using substitution method, let’s set u = tan(2x), then du = 2sec²(2x) dx. The integral becomes:

∫tan³(2x)sec(2x) dx = ∫(u³) du/2

This simplifies to:

(1/2) ∫u³ du

Integrating u³, we get:

(1/2) * (u⁴/4) + C

= u⁴/8 + C

Finally, substitute back u = tan(2x) to get the final result:

∫tan³(2x)sec(2x) dx = tan⁴(2x)/8 + C

16. tan⁴(x)

Solution:

Using the identity tan^2(x) = sec^2(x) – 1

= ∫(tan^2(x))^2 dx

= ∫(sec^2(x) – 1)^2 dx

= ∫(sec^4(x) – 2sec^2(x) + 1) dx

= 1/3 sec^3(x) – 2tan(x) + x + C

17. (sin³(x)+cos³(x))/sin²(x)cos²(x)

Solution:

To integrate the given expression, we can first simplify it using trigonometric identities:

[Tex]\frac{\sin^3(x) + \cos^3(x)}{\sin^2(x)\cos^2(x)} = \frac{(\sin(x) + \cos(x))(\sin^2(x) – \sin(x)\cos(x) + \cos^2(x))}{\sin^2(x)\cos^2(x)} \\[/Tex]

[Tex]= \frac{\sin(x) + \cos(x)}{\sin^2(x)} \\[/Tex]

[Tex]= \frac{\sin(x)}{\sin^2(x)} + \frac{\cos(x)}{\sin^2(x)}[/Tex]

[Tex]= \csc(x) + \cot(x)[/Tex]

Now, we can integrate the simplified expression:

[Tex]\int \frac{\sin^3(x) + \cos^3(x)}{\sin^2(x)\cos^2(x)} \,dx = \int (\csc(x) + \cot(x)) \,dx[/Tex]

[Tex]= -\ln|\csc(x) – \cot(x)| + C[/Tex]

[Tex]-\ln|\csc(x) – \cot(x)| + C[/Tex]

18. (cos(2x)+2sin²(x)/(cos²(x))

Solution:

Simplifying the integrand using trigonometric identities:

[Tex]\frac{\cos(2x) + 2\sin^2(x)}{\cos^2(x)} = \frac{\cos^2(x) – \sin^2(x) + 2\sin^2(x)}{\cos^2(x)} \\[/Tex]

[Tex]= \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} \\[/Tex]

[Tex]= \frac{1}{\cos^2(x)}[/Tex]

Now, we can perform the integration:

[Tex]\int \frac{\cos(2x) + 2\sin^2(x)}{\cos^2(x)} \,dx = \int \frac{1}{\cos^2(x)} \,dx \\[/Tex]

[Tex]= \int \sec^2(x) \,dx \\[/Tex]

[Tex]\tan(x) + C[/Tex]

19. [Tex]1/(\sin x \cos^3 x)[/Tex]

Solution:

Using substitution to simplify the integral, let u = cos(x), then du = -sin(x) dx and the integral becomes:

[Tex]\int \frac{1}{\sin(x)\cos^3(x)} \, dx = \int \frac{1}{(1 – u^2)u^3} \, du \\[/Tex]

[Tex]= \int \frac{1}{u^3 – u^5} \, du \\[/Tex]

[Tex]= \int \frac{1}{u^3(1 – u^2)} \, du \\[/Tex]

[Tex]= \int \left(\frac{A}{u} + \frac{B}{u^2} + \frac{C}{1 – u^2}\right) \, du[/Tex]

We can now find the values of A, B, and C by partial fraction decomposition:

[Tex]= A(u^2)(1 – u^2) + Bu(1 – u^2) + Cu(u^2) \\[/Tex]

[Tex]= A(u^2 – u^4) + Bu – Bu^3 + Cu^3 \\[/Tex]

[Tex]= A u^2 – A u^4 + Bu – Bu^3 + Cu^3 \\[/Tex]

[Tex]= (-A)u^4 + 0u^3 + (B – C)u^2 + (B)u[/Tex]

Equating coefficients:

1. A = -1

2. B = 0

3. B – C = 0 → C = B = 0

Thus, the integral becomes:

[Tex]\int \frac{1}{\sin(x)\cos^3(x)} \, dx = -\int \frac{1}{u} \, du \\[/Tex]

[Tex]= -\ln|u| + C \\[/Tex]

[Tex]-\ln|\cos(x)| + C[/Tex]

20. (cos2(x) /(cos(x)+sin(x))²

Solution:

Using trigonometric identity to simplify the integrand:

[Tex]\frac{\cos(2x)}{(\cos(x) + \sin(x))^2} = \frac{\cos^2(x) – \sin^2(x)}{(\cos(x) + \sin(x))^2}[/Tex]

Now, let’s rewrite the integrand in terms of cos(x) and sin(x) :

[Tex]\frac{\cos^2(x) – \sin^2(x)}{(\cos(x) + \sin(x))^2} = \frac{\cos^2(x) – \sin^2(x)}{\cos^2(x) + 2\sin(x)\cos(x) + \sin^2(x)}[/Tex]

Now, let [Tex]u = \sin(x), du = \cos(x) \, dx[/Tex]. The integral becomes:

[Tex]\int \frac{\cos^2(x) – \sin^2(x)}{\cos^2(x) + 2\sin(x)\cos(x) + \sin^2(x)} \, dx = \int \frac{\cos^2(x) – u^2}{\cos^2(x) + 2u\cos(x) + u^2} \, du \\[/Tex]

[Tex]= \int \frac{\cos^2(x) – u^2}{(\cos(x) + u)^2} \, du \\[/Tex]

[Tex]= \int \frac{\cos(x) – u}{\cos(x) + u} \, du \\[/Tex]

[Tex]= \int \frac{\cos(x)}{\cos(x) + u} \, du – \int \frac{u}{\cos(x) + u} \, du \\[/Tex]

[Tex]= \int \frac{\cos(x)}{\cos(x) + \sin(x)} \, du – \int \frac{\sin(x)}{\cos(x) + \sin(x)} \, du \\[/Tex]

[Tex]= \cos(x) \int \frac{1}{\cos(x) + \sin(x)} \, du – \sin(x) \int \frac{1}{\cos(x) + \sin(x)} \, du \\[/Tex]

[Tex]= \cos(x) \ln|\cos(x) + \sin(x)| – \sin(x) \ln|\cos(x) + \sin(x)| + C \\[/Tex]

[Tex](\cos(x) – \sin(x)) \ln|\cos(x) + \sin(x)| + C[/Tex]

21. sin^-1(cos(x))

Solution:

Let u = cos(x)

Then, du = -sin(x) dx

= -∫sin^{-1}(u) du

= -u sin^{-1}(u) + ∫du sin^{-1}(u)

= -u sin^{-1}(u) + u + C

= -cos(x) sin^{-1}(cos(x)) + cos(x) + C

22. 1/(cos(x-a)cos(x-b))

Solution:

= ∫(sec(x-a)sec(x-b)) dx

Let u = x – a, du = dx

= ∫sec(u)sec(u+b-a) du

= ∫sec(u)sec(u+b-a) du

= ∫sec^2(u)/(sec(b-a) + tan(b-a)tan(u)) du

= ∫du/(sec(b-a) + tan(b-a)tan(u))

= ∫du/(sec(b-a) + tan(b-a)tan(x-a))

= ∫du/(sec(b-a) + tan(b-a)u) + C

= ln|sec(b-a) + tan(b-a)u| + C

= ln|sec(b-a) + tan(b-a)(x-a)| + C

23. Solve ∫(sin²x – cos²x) / (sin²x cos²x) dx is equal to

(A) tan x + cot x + C

(B) tan x + cosec x + C

(C) – tan x + cot x + C

(D) tan x + sec x + C

Solution:

To solve the integral

[Tex]\int \frac{\sin^2 x – \cos^2 x}{\sin^2 x \cos^2 x} \, dx,[/Tex]

we can start by splitting the integrand into two separate fractions:

[Tex]\int \frac{\sin^2 x}{\sin^2 x \cos^2 x} \, dx – \int \frac{\cos^2 x}{\sin^2 x \cos^2 x} \, dx.[/Tex]

Simplify each fraction:

[Tex]\int \frac{1}{\cos^2 x} \, dx – \int \frac{1}{\sin^2 x} \, dx.[/Tex]

Now, using the basic integral formulas, we know that

[Tex]\int \frac{1}{\cos^2 x} \, dx = \tan x + C_1[/Tex]

and

[Tex]\int \frac{1}{\sin^2 x} \, dx = -\cot x + C_2.[/Tex]

Combining these, we get:

[Tex]\int \frac{\sin^2 x – \cos^2 x}{\sin^2 x \cos^2 x} \, dx = \tan x – \cot x + C.[/Tex]

So, the correct answer is (C) -tan x + cot x + C.\

24. ∫(e^x(1+x))/(cos²(e^xx)) dx equals to

(A) -cot(ex^x) + C

(B) tan(ex^x) + C

(C) tan(e^x) + C

(D) cot(e^x) + C

Solution:

To solve the integral [Tex]\int \frac{e^x(1+x)}{\cos^2(e^x x)} \, dx[/Tex], we can use the substitution method.

Let u = (e^x) x, then [Tex]du = (e^x + x e^x)[/Tex] , [Tex]dx = (1 + x)e^x \, dx[/Tex]. This transforms the integral into:

[Tex]\int \frac{1}{\cos^2(u)} \, du[/Tex]

which simplifies to:

[Tex]\int \sec^2(u) \, du = \tan(u) + C = \tan(e^x x) + C[/Tex]

So, the correct option is (B) tan(e^x x) + C.