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

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q20 – JEE-MAIN 2026

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Home Maths IntegralsQuestion #208

The value of the integral

\int_{\frac{π}{6}}^{\frac{π}{3}} (\frac{4 - \csc^{2} x}{\cos^{4} x}) d x

is:

Clue 1 of 3: Simplification Strategy

The integral appears complex, suggesting that the first step should be to simplify the integrand using trigonometric identities to break it down into functions with known antiderivatives.

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Ninja Strategy•Distractor Analysis

Recognize that options 2 and 4 correspond to common errors in applying the Fundamental Theorem of Calculus, which strongly suggests they are incorrect.

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Step 1: Simplify the Integrand✦ Active

Let the integral be $I$ . First, simplify the expression inside the integral. Rewrite $\csc^{2} x$ and use trigonometric identities.

\frac{4 - \csc^{2} x}{\cos^{4} x} = \frac{4 - \frac{1}{\sin^{2} x}}{\cos^{4} x} = \frac{4 \sin^{2} x - 1}{\sin^{2} x \cos^{4} x}

Now, use the identity $1 = \sin^{2} x + \cos^{2} x$ in the numerator:

\frac{4 \sin^{2} x - (\sin^{2} x + \cos^{2} x)}{\sin^{2} x \cos^{4} x} = \frac{3 \sin^{2} x - \cos^{2} x}{\sin^{2} x \cos^{4} x}

Split the fraction into two terms:

\frac{3 \sin^{2} x}{\sin^{2} x \cos^{4} x} - \frac{\cos^{2} x}{\sin^{2} x \cos^{4} x} = \frac{3}{\cos^{4} x} - \frac{1}{\sin^{2} x \cos^{2} x} = 3 \sec^{4} x - \frac{1}{(\sin x \cos x)^{2}}

Using the double angle identity $\sin (2 x) = 2 \sin x \cos x$ , the integrand becomes:

3 \sec^{4} x - \frac{1}{(\frac{1}{2} \sin (2 x))^{2}} = 3 \sec^{4} x - 4 \csc^{2} (2 x)

Step 2: Find the Antiderivative○ Expand

Now, integrate the simplified expression term by term. For the first term, write $\sec^{4} x = \sec^{2} x (1 + \tan^{2} x)$ and use substitution $u = \tan x$ .

\int 3 \sec^{4} x d x = \int 3 (1 + \tan^{2} x) \sec^{2} x d x = 3 (\tan x + \frac{\tan^{3} x}{3}) = 3 \tan x + \tan^{3} x

For the second term, use the standard integral of $\csc^{2} (a x)$ .

\int - 4 \csc^{2} (2 x) d x = - 4 (- \frac{\cot (2 x)}{2}) = 2 \cot (2 x)

The complete antiderivative $F (x)$ is:

F (x) = \tan^{3} x + 3 \tan x + 2 \cot (2 x)

Step 3: Apply the Fundamental Theorem of Calculus○ Expand

Evaluate the antiderivative at the upper and lower limits, $x = π / 3$ and $x = π / 6$ .

F (\frac{π}{3}) = \tan^{3} (\frac{π}{3}) + 3 \tan (\frac{π}{3}) + 2 \cot (\frac{2 π}{3}) = (\sqrt{3})^{3} + 3 (\sqrt{3}) + 2 (- \frac{1}{\sqrt{3}}) = 3 \sqrt{3} + 3 \sqrt{3} - \frac{2}{\sqrt{3}} = 6 \sqrt{3} - \frac{2}{\sqrt{3}} = \frac{18 - 2}{\sqrt{3}} = \frac{16}{\sqrt{3}}

F (\frac{π}{6}) = \tan^{3} (\frac{π}{6}) + 3 \tan (\frac{π}{6}) + 2 \cot (\frac{π}{3}) = {(\frac{1}{\sqrt{3}})}^{3} + 3 (\frac{1}{\sqrt{3}}) + 2 (\frac{1}{\sqrt{3}}) = \frac{1}{3 \sqrt{3}} + \frac{5}{\sqrt{3}} = \frac{1 + 15}{3 \sqrt{3}} = \frac{16}{3 \sqrt{3}}

The value of the integral is the difference:

I = F (\frac{π}{3}) - F (\frac{π}{6}) = \frac{16}{\sqrt{3}} - \frac{16}{3 \sqrt{3}} = \frac{48 - 16}{3 \sqrt{3}} = \frac{32}{3 \sqrt{3}}

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Maths Question 20 – JEE-MAIN 2026

Clue 1 of 3: Simplification Strategy

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