JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Definite integration using substitution with tangent and secant functions

Q: How did we know to integrate by parts instead of substituting tan x immediately?

Observing that the derivative of 1/(cos^4 x) generates (4sin x)/(cos^5 x) which multiplies with cot x to give 4/(cos^4 x), cancelling the other term directly.

The value of the integral integral from π/6 to π/3 of ((4 - cosec^2 x)/(cos^4 x)) dx is:

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A11/(√(3))B16/(√(3))C32/(3√(3))D64/(3√(3))

Step-by-step solutionView

Correct answer

The integral evaluates to $\frac{32}{3 \sqrt{3}}$ by expressing the integrand in terms of $\tan x$ and $\sec^{2} x$ and using the substitution $t = \tan x$ .

Option analysis

Why each option works or fails

A · 11/(√(3))

Evaluating the antiderivative $4 t + \frac{t^{3}}{3} + \frac{1}{t}$ with an arithmetic mistake when combining the fractional bounds. Substitute the bounds $t = \sqrt{3}$ and $t = \frac{1}{\sqrt{3}}$ carefully, ensuring all fractions share a common denominator before combining terms.

B · 16/(√(3))

Incorrectly rewriting $- \frac{{cosec}^{2} x}{\cos^{4} x}$ as $- \frac{\sec^{2} x}{\tan^{2} x}$ instead of $- \frac{\sec^{4} x}{\tan^{2} x}$ , losing a factor of $\sec^{2} x$ . Ensure that ${cosec}^{2} x / \cos^{4} x = \frac{1}{\sin^{2} x \cos^{4} x} = \frac{\sec^{4} x}{\tan^{2} x} = \frac{(1 + \tan^{2} x) \sec^{2} x}{\tan^{2} x}$ .

C · 32/(3√(3))

This is the correct value of the definite integral.

D · 64/(3√(3))

Adding the terms at the upper and lower limits instead of subtracting them, or doubling the result. By the Fundamental Theorem of Calculus, evaluate $F (b) - F (a)$ , taking care to subtract the lower limit value from the upper limit value.

Reviewed route

Solution

StepWorking

01Given

The integral to evaluate is $I = \int_{π / 6}^{π / 3} (\frac{4 - \csc^{2} x}{\cos^{4} x}) d x$ .

02Goal

Compute the exact value of the definite integral.

03Approach

Split the integral into two parts: $I_{1} = \int_{π / 6}^{π / 3} 4 \sec^{4} x d x$ and $I_{2} = \int_{π / 6}^{π / 3} \frac{\csc^{2} x}{\cos^{4} x} d x$ . Apply integration by parts on $I_{2}$ by taking $u = \cos^{- 4} x$ and $v^{'} = \csc^{2} x$ , which generates a term that cancels $I_{1}$ .

04Execute

Using integration by parts on $I_{2} = \int \frac{\csc^{2} x}{\cos^{4} x} d x$ : Let $u = \cos^{- 4} x ⟹ d u = - 4 \cos^{- 5} x (- \sin x) d x = 4 \frac{\sin x}{\cos^{5} x} d x$ . Let $d v = \csc^{2} x d x ⟹ v = - \cot x = - \frac{\cos x}{\sin x}$ . Then $I_{2} = {[- \frac{\cot x}{\cos^{4} x}]}_{π / 6}^{π / 3} - \int_{π / 6}^{π / 3} (- \frac{\cos x}{\sin x}) (4 \frac{\sin x}{\cos^{5} x}) d x = {[- \frac{\cot x}{\cos^{4} x}]}_{π / 6}^{π / 3} + \int_{π / 6}^{π / 3} \frac{4}{\cos^{4} x} d x$ . Therefore, $I = I_{1} - I_{2} = {[\frac{\cot x}{\cos^{4} x}]}_{π / 6}^{π / 3}$ .

05Execute

Evaluate the boundary term at limits $π / 3$ and $π / 6$ : At $x = π / 3$ : $\frac{\cot (π / 3)}{\cos^{4} (π / 3)} = \frac{1 / \sqrt{3}}{(1 / 2)^{4}} = \frac{16}{\sqrt{3}}$ . At $x = π / 6$ : $\frac{\cot (π / 6)}{\cos^{4} (π / 6)} = \frac{\sqrt{3}}{(\sqrt{3} / 2)^{4}} = \frac{\sqrt{3} \cdot 16}{9} = \frac{16}{3 \sqrt{3}}$ . Subtracting: $\frac{16}{\sqrt{3}} - \frac{16}{3 \sqrt{3}} = \frac{16}{\sqrt{3}} (1 - \frac{1}{3}) = \frac{32}{3 \sqrt{3}}$ .

✓Verify

Alternative check by converting to $\tan x$ : $\frac{4 - \csc^{2} x}{\cos^{4} x} = (4 - (1 + \cot^{2} x)) \sec^{4} x = (3 - \tan^{- 2} x) (1 + \tan^{2} x) \sec^{2} x$ . Let $t = \tan x$ : $\int_{1 / \sqrt{3}}^{\sqrt{3}} (3 + 3 t^{2} - t^{- 2} - 1) d t = {[2 t + t^{3} + t^{- 1}]}_{1 / \sqrt{3}}^{\sqrt{3}} = (2 \sqrt{3} + 3 \sqrt{3} + 1 / \sqrt{3}) - (2 / \sqrt{3} + 1 / (3 \sqrt{3}) + \sqrt{3}) = \frac{16}{\sqrt{3}} - \frac{16}{3 \sqrt{3}} = \frac{32}{3 \sqrt{3}}$ . Verified.

Hints that build this answer step by step

How can the integrand $\frac{4 - {cosec}^{2} x}{\cos^{4} x}$ be expressed in terms of $\tan x$ and $\sec^{2} x$ to prepare for the substitution $t = \tan x$ ?
$(4 (1 + \tan^{2} x) - \frac{1 + \tan^{2} x}{\tan^{2} x}) \sec^{2} x$
With the substitution $t = \tan x$ , where $d t = \sec^{2} x d x$ , what are the new limits and the simplified integrand in terms of $t$ ?
Limits from $\frac{1}{\sqrt{3}}$ to $\sqrt{3}$ , with integrand $3 + 4 t^{2} - t^{- 2}$
What is the value of ${[3 t + \frac{4 t^{3}}{3} + \frac{1}{t}]}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}$ ?
$\frac{32}{3 \sqrt{3}}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

How did we know to integrate by parts instead of substituting

\tan x

immediately?

Observing that the derivative of $\frac{1}{\cos^{4} x}$ generates $\frac{4 \sin x}{\cos^{5} x}$ which multiplies with $\cot x$ to give $\frac{4}{\cos^{4} x}$ , cancelling the other term directly.

Answer

The integral evaluates to $\frac{32}{3 \sqrt{3}}$ by expressing the integrand in terms of $\tan x$ and $\sec^{2} x$ and using the substitution $t = \tan x$ .

Why each option works or fails

A: 11/(√(3)) - Evaluating the antiderivative $4 t + \frac{t^{3}}{3} + \frac{1}{t}$ with an arithmetic mistake when combining the fractional bounds. Substitute the bounds $t = \sqrt{3}$ and $t = \frac{1}{\sqrt{3}}$ carefully, ensuring all fractions share a common denominator before combining terms.
B: 16/(√(3)) - Incorrectly rewriting $- \frac{{cosec}^{2} x}{\cos^{4} x}$ as $- \frac{\sec^{2} x}{\tan^{2} x}$ instead of $- \frac{\sec^{4} x}{\tan^{2} x}$ , losing a factor of $\sec^{2} x$ . Ensure that ${cosec}^{2} x / \cos^{4} x = \frac{1}{\sin^{2} x \cos^{4} x} = \frac{\sec^{4} x}{\tan^{2} x} = \frac{(1 + \tan^{2} x) \sec^{2} x}{\tan^{2} x}$ .
C · correct: 32/(3√(3)) - This is the correct value of the definite integral.
D: 64/(3√(3)) - Adding the terms at the upper and lower limits instead of subtracting them, or doubling the result. By the Fundamental Theorem of Calculus, evaluate $F (b) - F (a)$ , taking care to subtract the lower limit value from the upper limit value.

Step-by-step solution

Given: The integral to evaluate is $I = \int_{π / 6}^{π / 3} (\frac{4 - \csc^{2} x}{\cos^{4} x}) d x$ .
Goal: Compute the exact value of the definite integral.
Approach: Split the integral into two parts: $I_{1} = \int_{π / 6}^{π / 3} 4 \sec^{4} x d x$ and $I_{2} = \int_{π / 6}^{π / 3} \frac{\csc^{2} x}{\cos^{4} x} d x$ . Apply integration by parts on $I_{2}$ by taking $u = \cos^{- 4} x$ and $v^{'} = \csc^{2} x$ , which generates a term that cancels $I_{1}$ .
Execute: Using integration by parts on $I_{2} = \int \frac{\csc^{2} x}{\cos^{4} x} d x$ : Let $u = \cos^{- 4} x ⟹ d u = - 4 \cos^{- 5} x (- \sin x) d x = 4 \frac{\sin x}{\cos^{5} x} d x$ . Let $d v = \csc^{2} x d x ⟹ v = - \cot x = - \frac{\cos x}{\sin x}$ . Then $I_{2} = {[- \frac{\cot x}{\cos^{4} x}]}_{π / 6}^{π / 3} - \int_{π / 6}^{π / 3} (- \frac{\cos x}{\sin x}) (4 \frac{\sin x}{\cos^{5} x}) d x = {[- \frac{\cot x}{\cos^{4} x}]}_{π / 6}^{π / 3} + \int_{π / 6}^{π / 3} \frac{4}{\cos^{4} x} d x$ . Therefore, $I = I_{1} - I_{2} = {[\frac{\cot x}{\cos^{4} x}]}_{π / 6}^{π / 3}$ .
Execute: Evaluate the boundary term at limits $π / 3$ and $π / 6$ : At $x = π / 3$ : $\frac{\cot (π / 3)}{\cos^{4} (π / 3)} = \frac{1 / \sqrt{3}}{(1 / 2)^{4}} = \frac{16}{\sqrt{3}}$ . At $x = π / 6$ : $\frac{\cot (π / 6)}{\cos^{4} (π / 6)} = \frac{\sqrt{3}}{(\sqrt{3} / 2)^{4}} = \frac{\sqrt{3} \cdot 16}{9} = \frac{16}{3 \sqrt{3}}$ . Subtracting: $\frac{16}{\sqrt{3}} - \frac{16}{3 \sqrt{3}} = \frac{16}{\sqrt{3}} (1 - \frac{1}{3}) = \frac{32}{3 \sqrt{3}}$ .
Verify: Alternative check by converting to $\tan x$ : $\frac{4 - \csc^{2} x}{\cos^{4} x} = (4 - (1 + \cot^{2} x)) \sec^{4} x = (3 - \tan^{- 2} x) (1 + \tan^{2} x) \sec^{2} x$ . Let $t = \tan x$ : $\int_{1 / \sqrt{3}}^{\sqrt{3}} (3 + 3 t^{2} - t^{- 2} - 1) d t = {[2 t + t^{3} + t^{- 1}]}_{1 / \sqrt{3}}^{\sqrt{3}} = (2 \sqrt{3} + 3 \sqrt{3} + 1 / \sqrt{3}) - (2 / \sqrt{3} + 1 / (3 \sqrt{3}) + \sqrt{3}) = \frac{16}{\sqrt{3}} - \frac{16}{3 \sqrt{3}} = \frac{32}{3 \sqrt{3}}$ . Verified.

Definite integration using substitution with tangent and secant functions

What feels right?

The integral evaluates to $\frac{32}{3 \sqrt{3}}$ by expressing the integrand in terms of $\tan x$ and $\sec^{2} x$ and using the substitution $t = \tan x$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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