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Integral Calculus: JEE Main Mathematics Question with Solution

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

Observing that the derivative of generates which multiplies with to give , cancelling the other term directly.

The value of the integral

\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \frac{4 - \operatorname{cosec}^2 x}{\cos^4 x} \right) dx

is :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{11}{\sqrt{3}}

\frac{16}{\sqrt{3}}

\frac{32}{3\sqrt{3}}

\frac{64}{3\sqrt{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 ·

\frac{11}{\sqrt{3}}

Evaluating the antiderivative $4t + \frac{t^3}{3} + \frac{1}{t}$ 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 ·

\frac{16}{\sqrt{3}}

Incorrectly rewriting $-\frac{\operatorname{cosec}^2 x}{\cos^4 x}$ 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 $\operatorname{cosec}^2 x / \cos^4 x = \frac{1}{\sin^2 x \cos^4 x} = \frac{\sec^4 x}{\tan^2 x} = \frac{(1 + \tan^2 x)\sec^2 x}{\tan^2 x}$ .

C ·

\frac{32}{3\sqrt{3}}

None. This is the correct value of the definite integral. None.

D ·

\frac{64}{3\sqrt{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 $I = \int_{\pi/6}^{\pi/3} \left( \frac{4 - \csc^2 x}{\cos^4 x} \right) dx$ .

02goal

Compute the exact value of the definite integral.

03approach

Split the integral into two parts: $I_1 = \int_{\pi/6}^{\pi/3} 4\sec^4 x\,dx$ and $I_2 = \int_{\pi/6}^{\pi/3} \frac{\csc^2 x}{\cos^4 x}\,dx$ . 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 $I_2 = \int \frac{\csc^2 x}{\cos^4 x} dx$ : Let $u = \cos^{-4} x \implies du = -4\cos^{-5} x(-\sin x)dx = 4\frac{\sin x}{\cos^5 x} dx$ . Let $dv = \csc^2 x dx \implies v = -\cot x = -\frac{\cos x}{\sin x}$ . Then $I_2 = \left[ -\frac{\cot x}{\cos^4 x} \right]_{\pi/6}^{\pi/3} - \int_{\pi/6}^{\pi/3} \left(-\frac{\cos x}{\sin x}\right) \left(4\frac{\sin x}{\cos^5 x}\right) dx = \left[ -\frac{\cot x}{\cos^4 x} \right]_{\pi/6}^{\pi/3} + \int_{\pi/6}^{\pi/3} \frac{4}{\cos^4 x} dx$ . Therefore, $I = I_1 - I_2 = \left[ \frac{\cot x}{\cos^4 x} \right]_{\pi/6}^{\pi/3}$ .

05execute

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

✓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 $t = \tan x$ : $\int_{1/\sqrt{3}}^{\sqrt{3}} (3 + 3t^2 - t^{-2} - 1) dt = \left[ 2t + t^3 + t^{-1} \right]_{1/\sqrt{3}}^{\sqrt{3}} = (2\sqrt{3} + 3\sqrt{3} + 1/\sqrt{3}) - (2/\sqrt{3} + 1/(3\sqrt{3}) + \sqrt{3}) = \frac{16}{\sqrt{3}} - \frac{16}{3\sqrt{3}} = \frac{32}{3\sqrt{3}}$ . Verified.

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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: $\frac{11}{\sqrt{3}}$ - Evaluating the antiderivative $4t + \frac{t^3}{3} + \frac{1}{t}$ 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: $\frac{16}{\sqrt{3}}$ - Incorrectly rewriting $-\frac{\operatorname{cosec}^2 x}{\cos^4 x}$ 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 $\operatorname{cosec}^2 x / \cos^4 x = \frac{1}{\sin^2 x \cos^4 x} = \frac{\sec^4 x}{\tan^2 x} = \frac{(1 + \tan^2 x)\sec^2 x}{\tan^2 x}$ .
C · correct: $\frac{32}{3\sqrt{3}}$ - None. This is the correct value of the definite integral. None.
D: $\frac{64}{3\sqrt{3}}$ - 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 $I = \int_{\pi/6}^{\pi/3} \left( \frac{4 - \csc^2 x}{\cos^4 x} \right) dx$ .
goal: Compute the exact value of the definite integral.
approach: Split the integral into two parts: $I_1 = \int_{\pi/6}^{\pi/3} 4\sec^4 x\,dx$ and $I_2 = \int_{\pi/6}^{\pi/3} \frac{\csc^2 x}{\cos^4 x}\,dx$ . 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 $I_2 = \int \frac{\csc^2 x}{\cos^4 x} dx$ : Let $u = \cos^{-4} x \implies du = -4\cos^{-5} x(-\sin x)dx = 4\frac{\sin x}{\cos^5 x} dx$ . Let $dv = \csc^2 x dx \implies v = -\cot x = -\frac{\cos x}{\sin x}$ . Then $I_2 = \left[ -\frac{\cot x}{\cos^4 x} \right]_{\pi/6}^{\pi/3} - \int_{\pi/6}^{\pi/3} \left(-\frac{\cos x}{\sin x}\right) \left(4\frac{\sin x}{\cos^5 x}\right) dx = \left[ -\frac{\cot x}{\cos^4 x} \right]_{\pi/6}^{\pi/3} + \int_{\pi/6}^{\pi/3} \frac{4}{\cos^4 x} dx$ . Therefore, $I = I_1 - I_2 = \left[ \frac{\cot x}{\cos^4 x} \right]_{\pi/6}^{\pi/3}$ .
execute: Evaluate the boundary term at limits $\pi/3$ and $\pi/6$ : At $x = \pi/3$ : $\frac{\cot(\pi/3)}{\cos^4(\pi/3)} = \frac{1/\sqrt{3}}{(1/2)^4} = \frac{16}{\sqrt{3}}$ . At $x = \pi/6$ : $\frac{\cot(\pi/6)}{\cos^4(\pi/6)} = \frac{\sqrt{3}}{(\sqrt{3}/2)^4} = \frac{\sqrt{3} \cdot 16}{9} = \frac{16}{3\sqrt{3}}$ . Subtracting: $\frac{16}{\sqrt{3}} - \frac{16}{3\sqrt{3}} = \frac{16}{\sqrt{3}}\left(1 - \frac{1}{3}\right) = \frac{32}{3\sqrt{3}}$ .
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 $t = \tan x$ : $\int_{1/\sqrt{3}}^{\sqrt{3}} (3 + 3t^2 - t^{-2} - 1) dt = \left[ 2t + t^3 + t^{-1} \right]_{1/\sqrt{3}}^{\sqrt{3}} = (2\sqrt{3} + 3\sqrt{3} + 1/\sqrt{3}) - (2/\sqrt{3} + 1/(3\sqrt{3}) + \sqrt{3}) = \frac{16}{\sqrt{3}} - \frac{16}{3\sqrt{3}} = \frac{32}{3\sqrt{3}}$ . Verified.

Integral Calculus: JEE Main Mathematics Question with Solution

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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