Integration by Substitution Maths12: Imp Questions 2024

Introduction

Among the various chapters, Integration by Substitution Maths12 is an essential topic that not only tests a student’s understanding of integrals but also their problem-solving skills. Class 12 is a crucial year in a student’s academic life, and Mathematics plays a significant role in securing high scores. In this blog post, we will cover some of the most important questions from this chapter along with step-by-step solutions, which will help you strengthen your concepts and excel in your exams.

What is Integration by Substitution Maths12?

Integration by substitution is a method used to simplify complex integrals by making a substitution to change the variable of integration. Integration by Substitution Maths12 is often compared to the reverse process of the chain rule in differentiation. This technique helps in transforming integrals into simpler forms that are easier to evaluate.

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Important Formulae for Integration by Substitution Maths12

Before diving into the questions, here are some important formulae that you will need:

• ∫ f(g(x)) g'(x) dx = ∫ f(u) du
• ∫ e^x dx = e^x + C
• ∫ (1/x) dx = ln|x| + C
• ∫ sin(x) dx = -cos(x) + C
• ∫ cos(x) dx = sin(x) + C

Important Questions with Solutions: Integration by Substitution Maths12

Question 1:

Evaluate the integral:
∫ (2x) / (x² + 1) dx

Solution:

Step 1: Let’s make the substitution. Let

$u = x^2 + 1$

, so

$du/dx = 2x$

Step 2: Rewrite the integral using the substitution. Now,

$dx = du / 2x$

so the integral becomes:

$∫ \frac{2x}{u} \times \frac{du}{2x} = ∫ \frac{1}{u} du$

Step 3: Integrate the simplified expression.

$∫ \frac{1}{u} du = ln|u| + C$

Step 4: Substitute

$u = x^2 + 1$

back into the result.

$ln|x^2 + 1| + C$

Thus, the solution is

$ln|x^2 + 1| + C$

 

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Question 2:

Evaluate the integral:
∫ cos(2x) dx

Solution:

Step 1: Make the substitution. Let

$u = 2x$

, so

$du/dx = 2$

or

$dx = du / 2$

Step 2: Rewrite the integral using this substitution.

$∫ cos(2x) dx = \frac{1}{2} ∫ cos(u) du$

Step 3: Integrate the simplified expression.

$\frac{1}{2} ∫ cos(u) du = \frac{1}{2} sin(u) + C$

Step 4: Substitute

$u = 2x$

back into the result.

$\frac{1}{2} sin(2x) + C$

Thus, the solution is

$\frac{1}{2} sin(2x) + C$

Question 3:

Evaluate the integral:
∫ e^(3x) dx

Solution:

Step 1: Make the substitution. Let

$u = 3x$

, so

$du/dx = 3$

or

$dx = du / 3$

Step 2: Rewrite the integral.

$∫ e^(3x) dx = \frac{1}{3} ∫ e^u du$

Step 3: Integrate.

$\frac{1}{3} e^u + C$

Step 4: Substitute

$u = 3x$

back into the result.

$\frac{1}{3} e^(3x) + C$

Thus, the solution is

$\frac{1}{3} e^(3x) + C$

 

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Question 4:

Evaluate the integral:
∫ sin(x) cos(x) dx

Solution:

Step 1: Use the substitution

$u = sin(x)$

, so

$du/dx = cos(x)$

Step 2: Rewrite the integral.

$∫ sin(x) cos(x) dx = ∫ u du$

Step 3: Integrate the simplified expression.

$∫ u du = \frac{u^2}{2} + C$

Step 4: Substitute

$u = sin(x)$

back into the result.

$\frac{sin^2(x)}{2} + C$

Thus, the solution is

$\frac{sin^2(x)}{2} + C$

Question 5:

Evaluate the integral:
∫ (3x² + 2x) e^(x³ + x²) dx

Solution:

Step 1: Let

$u = x^3 + x^2$

, so

$du/dx = 3x^2 + 2x$

Step 2: Rewrite the integral:

$∫ (3x^2 + 2x) e^{u} dx = ∫ e^{u} du$

Step 3: Integrate:

$e^u + C$

Step 4: Substitute back

$u = x^3 + x^2$

Thus, the solution is

$e^{x^3 + x^2} + C$

 

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Question 6:

Evaluate the integral:
∫ (x / (x² + 4)) dx

Solution:

Step 1: Let

$u = x^2 + 4$

, so

$du/dx = 2x$

or

$dx = du / 2x$

 

Step 2: Rewrite the integral:

$\frac{1}{2} ∫ \frac{du}{u}$

​Step 3: Integrate:

$\frac{1}{2} ln|u| + C$

Step 4: Substitute back

$u = x^2 + 4$

Thus, the solution is

$\frac{1}{2} ln|x^2 + 4| + C$

 

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Question 7:

Evaluate the integral:
∫ x sin(x²) dx

Solution:

Step 1: Let

$u = x^2$

, so

$du/dx = 2x$

or

$dx = du / 2x$

Step 2: Rewrite the integral:

$\frac{1}{2} ∫ sin(u) du$

Step 3: Integrate:

$\frac{1}{2} (-cos(u)) + C$

Step 4: Substitute back

$u = x^2$

:Thus, the solution is

$-\frac{1}{2} cos(x^2) + C$

 

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Question 8:

Evaluate the integral:
∫ (2x / (x² + 9)) dx

Solution:

Step 1: Let

$u = x^2 + 9$

, so

$du/dx = 2x$

Step 2: Rewrite the integral:

$∫ \frac{1}{u} du$

Step 3: Integrate:

$ln|u| + C$

Step 4: Substitute back

$u = x^2 + 9$

:Thus, the solution is

$ln|x^2 + 9| + C$

 

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Question 9:

Evaluate the integral:
∫ sin(3x) dx

Solution:

Step 1: Let

$u = 3x$

u=3x, so

$du/dx = 3$

or

$dx = du / 3$

Step 2: Rewrite the integral:

$\frac{1}{3} ∫ sin(u) du$

Step 3: Integrate:

$\frac{1}{3} (-cos(u)) + C$

Step 4: Substitute back

$u = 3x$

 

$-\frac{1}{3} cos(3x) + C$

Thus, the solution is

$-\frac{1}{3} cos(3x) + C$

 

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Question 10:

Evaluate the integral:
∫ x³ cos(x⁴) dx

Solution:

Step 1: Let

$u = x^4$

, so

$du/dx = 4x^3$

or

$dx = du / 4x^3$

Step 2: Rewrite the integral:

$\frac{1}{4} ∫ cos(u) du$

Step 3: Integrate:

$\frac{1}{4} sin(u) + C$

Step 4: Substitute back

$u = x^4$ $\frac{1}{4} sin(x^4) + C$

Thus, the solution is

$\frac{1}{4} sin(x^4) + C$

 

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Question 11:

Evaluate the integral:
∫ (1 / (x² + 16)) dx

Solution:

Step 1: Let

$u = x^2 + 16$

, so

$du/dx = 2x$

Step 2: Split the integral:

$\frac{1}{2} ∫ \frac{1}{u} du$

Step 3: Integrate:

$\frac{1}{2} ln|u| + C$

Step 4: Substitute back

$u = x^2 + 16$

 

$\frac{1}{2} ln|x^2 + 16| + C$

Thus, the solution is

$\frac{1}{2} ln|x^2 + 16| + C$

 

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Question 12:

Evaluate the integral:
∫ e^(5x) dx

Solution:

Step 1: Let

$u = 5x$

, so

$du/dx = 5$

or

$dx = du / 5$

Step 2: Rewrite the integral:

$\frac{1}{5} ∫ e^u du$

Step 3: Integrate:

$\frac{1}{5} e^u + C$

Step 4: Substitute back

$u = 5x$ $\frac{1}{5} e^{5x} + C$

Thus, the solution is

$\frac{1}{5} e^{5x} + C$

 

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Question 13:

Evaluate the integral:
∫ (x / sqrt(x² + 1)) dx

Solution:

Step 1: Let

$u = x^2 + 1$

, so

$du/dx = 2x$

Step 2: Rewrite the integral:

$\frac{1}{2} ∫ \frac{du}{\sqrt{u}}$

​Step 3: Integrate:

$\frac{1}{2} \times 2 \sqrt{u} + C$

Step 4: Substitute back

$u = x^2 + 1$

 

$\sqrt{x^2 + 1} + C$

Thus, the solution is

$\sqrt{x^2 + 1} + C$

 

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Question 14:

Evaluate the integral:
∫ tan(x) dx

Solution:

Step 1: Use the identity:

$tan(x) = \frac{sin(x)}{cos(x)}$

​Step 2: Let

$u = cos(x)$

, so

$du/dx = -sin(x)$

Step 3: Rewrite the integral:

$– ∫ \frac{du}{u}$

​Step 4: Integrate:

$-ln|u| + C$

Step 5: Substitute back

$u = cos(x)$ $-ln|cos(x)| + C$

hus, the solution is

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

 

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Question 15:

Evaluate the integral:
∫ sec²(3x) dx

Solution:

Step 1: Let

$u = 3x$

, so

$du/dx = 3$

or

$dx = du / 3$

Step 2: Rewrite the integral:

$\frac{1}{3} ∫ sec^2(u) du$

Step 3: Integrate:

$\frac{1}{3} tan(u) + C$

Step 4: Substitute back

$u = 3x$ $\frac{1}{3} tan(3x) + C$

Thus, the solution is

$\frac{1}{3} tan(3x) + C$

.

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Question 16:

Evaluate the integral:
∫ (x² / (x³ + 5)) dx

Solution:

Step 1: Let

$u = x^3 + 5$

, so

$du/dx = 3x^2$

or

$dx = du / 3x^2$

Step 2: Rewrite the integral:

$\frac{1}{3} ∫ \frac{du}{u}$

​Step 3: Integrate:

$\frac{1}{3} ln|u| + C$

Step 4: Substitute back

$u = x^3 + 5$ $\frac{1}{3} ln|x^3 + 5| + C$

Thus, the solution is

$\frac{1}{3} ln|x^3 + 5| + C$

 

Question 17:

Evaluate the integral:
∫ sin²(x) dx

Solution:

Step 1: Use the identity:

$sin^2(x) = \frac{1 – cos(2x)}{2}$

​Step 2: Rewrite the integral:

$\frac{1}{2} ∫ (1 – cos(2x)) dx$

Step 3: Integrate:

$\frac{1}{2} [x – \frac{sin(2x)}{2}] + C$

Thus, the solution is

$\frac{x}{2} – \frac{sin(2x)}{4} + C$

 

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Question 18:

Evaluate the integral:
∫ cos²(x) dx

Solution:

Step 1: Use the identity:

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

​Step 2: Rewrite the integral:

$\frac{1}{2} ∫ (1 + cos(2x)) dx$

Step 3: Integrate:

$\frac{1}{2} [x + \frac{sin(2x)}{2}] + C$

Thus, the solution is

$\frac{x}{2} + \frac{sin(2x)}{4} + C$

 

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Question 19:

Evaluate the integral:
∫ 1 / sqrt(1 – x²) dx

Solution:

Step 1: Let

$u = sin^{-1}(x)$

, so

$du/dx = 1/\sqrt{1 – x^2}$

Step 2: The integral becomes:

$∫ du = u + C$

Step 3: Substitute back

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

Thus, the solution is

$sin^{-1}(x) + C$

 

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Question 20:

Evaluate the integral:
∫ x / sqrt(x² + 4) dx

Solution:

Step 1: Let

$u = x^2 + 4$

, so

$du/dx = 2x$

or

$dx = du / 2x$

Step 2: Rewrite the integral:

$\frac{1}{2} ∫ \frac{du}{\sqrt{u}}$

​Step 3: Integrate:

$\sqrt{u} + C$

Step 4: Substitute back

$u = x^2 + 4$

 

$\sqrt{x^2 + 4} + C$

Thus, the solution is

$\sqrt{x^2 + 4} + C$

 

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Question 21:

Evaluate the integral:
∫ ln(x) / x dx

Solution:

Step 1: Let

$u = ln(x)$

, so

$du/dx = 1/x$

Step 2: Rewrite the integral:

$∫ u du$

Step 3: Integrate:

$\frac{u^2}{2} + C$

Step 4: Substitute back

$u = ln(x)$ $\frac{ln^2(x)}{2} + C$

Thus, the solution is

$\frac{ln^2(x)}{2} + C$

 

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Tips for Solving Integration by Substitution Maths12 Questions

Identify the inner function: Look for composite functions where substitution can be applied. Usually, the function inside parentheses or under a square root is a good candidate for substitution.

Match the differential: Ensure that after substitution, you can express the entire integral in terms of the new variable. Don’t forget to change:-

$d\mathrm{x\; to\; du.}$ $$

$$

Simplify the integral: After substitution, the integral should become simpler to solve. Apply the basic integration rules to proceed.

Don’t forget to back-substitute: Once you’ve integrated the expression, substitute back the original variable before writing the final answer.

Conclusion: Integration by Substitution Maths12

These 21 additional questions with detailed solutions will help you prepare efficiently for the topic of Integration by Substitution Maths12.

Make sure to practice these problems to strengthen your grasp on the subject and perform well in your exams.

https://pcmconceptclear.com/integration-basic-concept-and-formula/