Class 12 NCERT Mathematics Solutions - Chapter 7- Integrals - Exercise 7.3

Chapter 7 of the Class 12 NCERT Mathematics Part II textbook, titled "Integrals," covers the fundamental concepts of integral calculus. This chapter introduces techniques for finding integrals, including definite and indefinite integrals, and their applications. Exercise 7.3 focuses on solving problems related to various integration techniques and applications.

This section provides detailed solutions for Exercise 7.3 from Chapter 7 of the Class 12 NCERT Mathematics Part II textbook. The exercise includes problems that require applying integration techniques to find the integral of functions. Solutions are presented step-by-step to help students understand and apply integration methods effectively.

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

Class 12 NCERT Mathematics Solutions- Exercise 7.3

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 \cos^2(x) + \sin^2(x) = 1 to simplify the expression:

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)}

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

\int \frac{1 - \cos(x)}{1 + \cos(x)} \, dx = \int \frac{(1 - \cos(x))^2}{\sin^2(x)} \, dx

3. Next, we'll expand the numerator ((1 - \cos(x))^2:

(1 - \cos(x))^2 = 1 - 2\cos(x) + \cos^2(x)

4. Now, our integral becomes:

\int \frac{1 - \cos(x)}{1 + \cos(x)} \, dx = \int \frac{1 - 2\cos(x) + \cos^2(x)}{\sin^2(x)} \, dx

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

\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

6. Now, we'll integrate each term:

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

= \int \csc^2(x) \, dx = -\cot(x) + C_1

- \int \frac{\cos(x)}{\sin^2(x)} \, dx can be simplified using substitution:

Let u = \sin(x), then du = \cos(x) \, dx.

The integral becomes:

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

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

= \int \frac{1 - \sin^2(x)}{\sin^2(x)} \, dx = \int \left(\frac{1}{\sin^2(x)} - 1\right) \, dx

= -\cot(x) - x + C_3

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

-\cot(x) - \cot(x) - x + C = -2\cot(x) - x + C

So, \int \frac{1 - \cos(x)}{1 + \cos(x)} \, dx = -2\cot(x) - x + C, 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 (1 - \cos(x)) to get rid of the denominator:

\int \frac{\cos(x)}{1 + \cos(x)} \, dx = \int \frac{\cos(x)(1 - \cos(x))}{(1 + \cos(x))(1 - \cos(x))} \, dx

= \int \frac{\cos(x) - \cos^2(x)}{1 - \cos^2(x)} \, dx

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

1 - \cos^2(x) = \sin^2(x)

\int \frac{\cos(x) - \cos^2(x)}{\sin^2(x)} \, dx

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

\int \frac{\cos(x)}{\sin^2(x)} \, dx - \int \frac{\cos^2(x)}{\sin^2(x)} \, dx

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

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

5. For the second integral, we use the Pythagorean identity \cos^2(x) = 1 - \sin^2(x):

\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

= -\cot(x) - x + C_2

-\cot(x) + (-\cot(x) - x) + C = -2\cot(x) - x + C

\int \frac{\cos(x)}{1 + \cos(x)} \, dx = -2\cot(x) - x + C

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 \sin^2(x) = 1 - \cos^2(x) to rewrite the integrand:

\frac{\sin^2(x)}{1 + \cos(x)} = \frac{1 - \cos^2(x)}{1 + \cos(x)}

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

\int \frac{\sin^2(x)}{1 + \cos(x)} \, dx = \int \frac{1 - \cos^2(x)}{1 + \cos(x)} \, dx

3. Now, we'll simplify the expression further by dividing the numerator by 1 + \cos(x):

= \int \frac{1}{1 + \cos(x)} \, dx - \int \frac{\cos^2(x)}{1 + \cos(x)} \, dx

4. We can now integrate each term separately:

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

= \int \frac{1}{1 + \cos(x)} \, dx

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

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

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

= 2\tan\left(\frac{x}{2}\right) + C_1

For the second integral, we'll use a substitution u = \sin(x), du = \cos(x) \, dx:

= \int \frac{1 - u^2}{1 + u} \, du

= \int \frac{1}{1 + u} \, du - \int u \, du

= \ln|1 + u| - \frac{u^2}{2} + C_2

2\tan\left(\frac{x}{2}\right) + \ln|1 + \sin(x)| - \frac{\sin^2(x)}{2} + C

\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

13. (cos(2x)-cos(2α)/cos(x)cos(α))

Solution:

1. We'll start by using the double angle formula for cosine: \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta).

\cos(2x) - \cos(2\alpha) = (\cos^2(x) - \sin^2(x)) - (\cos^2(\alpha) - \sin^2(\alpha))

= \cos^2(x) - \cos^2(\alpha) - (\sin^2(x) - \sin^2(\alpha))

2. Next, we'll use the identity \sin^2(\theta) = 1 - \cos^2(\theta) to rewrite the expression:

\cos(2x) - \cos(2\alpha) = \cos^2(x) - \cos^2(\alpha) - (1 - \cos^2(x) - (1 - \cos^2(\alpha)))

= \cos^2(x) - \cos^2(\alpha) - 1 + \cos^2(x) + 1 - \cos^2(\alpha)

= 2\cos^2(x) - 2\cos^2(\alpha)

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

\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

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

= 2\int \frac{\cos^2(x) - \cos^2(\alpha)}{\cos(x) - \cos(\alpha)} \, dx

5. Now, we'll use the difference of squares identity: \cos^2(x) - \cos^2(\alpha) = (\cos(x) + \cos(\alpha))(\cos(x) - \cos(\alpha))

= 2\int \frac{(\cos(x) + \cos(\alpha))(\cos(x) - \cos(\alpha))}{\cos(x) - \cos(\alpha)} \, dx

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

= 2\int \cos(x) + \cos(\alpha) \, dx

7. Now, we can integrate each term separately:

= 2\left(\int \cos(x) \, dx + \int \cos(\alpha) \, dx\right)

= 2(\sin(x) + \cos(\alpha)x) + C

\int \frac{\cos(2x) - \cos(2\alpha)}{\cos(x) - \cos(\alpha)} \, dx = 2(\sin(x) + \cos(\alpha)x) + C

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:

\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)} \\

= \frac{\sin(x) + \cos(x)}{\sin^2(x)} \\

= \frac{\sin(x)}{\sin^2(x)} + \frac{\cos(x)}{\sin^2(x)}

= \csc(x) + \cot(x)

Now, we can integrate the simplified expression:

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

= -\ln|\csc(x) - \cot(x)| + C

-\ln|\csc(x) - \cot(x)| + C

18. (cos(2x)+2sin²(x)/(cos²(x))

Solution:

Simplifying the integrand using trigonometric identities:

\frac{\cos(2x) + 2\sin^2(x)}{\cos^2(x)} = \frac{\cos^2(x) - \sin^2(x) + 2\sin^2(x)}{\cos^2(x)} \\

= \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} \\

= \frac{1}{\cos^2(x)}

Now, we can perform the integration:

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

= \int \sec^2(x) \,dx \\

\tan(x) + C

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

Solution:

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

\int \frac{1}{\sin(x)\cos^3(x)} \, dx = \int \frac{1}{(1 - u^2)u^3} \, du \\

= \int \frac{1}{u^3 - u^5} \, du \\

= \int \frac{1}{u^3(1 - u^2)} \, du \\

= \int \left(\frac{A}{u} + \frac{B}{u^2} + \frac{C}{1 - u^2}\right) \, du

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

= A(u^2)(1 - u^2) + Bu(1 - u^2) + Cu(u^2) \\

= A(u^2 - u^4) + Bu - Bu^3 + Cu^3 \\

= A u^2 - A u^4 + Bu - Bu^3 + Cu^3 \\

= (-A)u^4 + 0u^3 + (B - C)u^2 + (B)u

Equating coefficients:

1. A = -1

2. B = 0

3. B - C = 0 → C = B = 0

Thus, the integral becomes:

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

= -\ln|u| + C \\

-\ln|\cos(x)| + C

20. (cos2(x) /(cos(x)+sin(x))²

Solution:

Using trigonometric identity to simplify the integrand:

\frac{\cos(2x)}{(\cos(x) + \sin(x))^2} = \frac{\cos^2(x) - \sin^2(x)}{(\cos(x) + \sin(x))^2}

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

\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)}

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

\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 \\

= \int \frac{\cos^2(x) - u^2}{(\cos(x) + u)^2} \, du \\

= \int \frac{\cos(x) - u}{\cos(x) + u} \, du \\

= \int \frac{\cos(x)}{\cos(x) + u} \, du - \int \frac{u}{\cos(x) + u} \, du \\

= \int \frac{\cos(x)}{\cos(x) + \sin(x)} \, du - \int \frac{\sin(x)}{\cos(x) + \sin(x)} \, du \\

= \cos(x) \int \frac{1}{\cos(x) + \sin(x)} \, du - \sin(x) \int \frac{1}{\cos(x) + \sin(x)} \, du \\

= \cos(x) \ln|\cos(x) + \sin(x)| - \sin(x) \ln|\cos(x) + \sin(x)| + C \\

(\cos(x) - \sin(x)) \ln|\cos(x) + \sin(x)| + C

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

\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} \, dx,

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

\int \frac{\sin^2 x}{\sin^2 x \cos^2 x} \, dx - \int \frac{\cos^2 x}{\sin^2 x \cos^2 x} \, dx.

Simplify each fraction:

\int \frac{1}{\cos^2 x} \, dx - \int \frac{1}{\sin^2 x} \, dx.

Now, using the basic integral formulas, we know that

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

and

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

Combining these, we get:

\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} \, dx = \tan x - \cot x + C.

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 \int \frac{e^x(1+x)}{\cos^2(e^x x)} \, dx, we can use the substitution method.

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

\int \frac{1}{\cos^2(u)} \, du

which simplifies to:

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

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

Related Articles,

• Integrals
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.2
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.4

Conclusion

Chapter 7 of the Class 12 NCERT Mathematics Part II textbook, "Integrals," focuses on techniques for finding and applying integrals. Exercise 7.3 involves solving problems related to various integration methods. Solutions are provided step-by-step to help students understand and apply integral calculus effectively.