If 1-5 ^2 x ^5 x ^2 x dx=f(x)+C, where C is the constant of integ… - Solution | JEE Mathematics

MCQMediumJEE 2026Indefinite Integrals

If $\int \frac{1-5\cos^2 x}{\sin^5 x\cos^2 x}\,dx=f(x)+C,$ where $C$ is the constant of integration, then $f\!\left(\frac{\pi}{6}\right)-f\!\left(\frac{\pi}{4}\right)$ is equal to:

A
$\dfrac{1}{\sqrt{3}}(26-\sqrt{3})$
B
$\dfrac{1}{\sqrt{3}}(26+\sqrt{3})$
C
$\dfrac{4}{\sqrt{3}}(8-\sqrt{6})$
D
$\dfrac{2}{\sqrt{3}}(4+\sqrt{6})$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

\int \frac{1-5\cos^2 x}{\sin^5 x\cos^2 x}\,dx=f(x)+C

Find: $f\!\left(\frac{\pi}{6}\right)-f\!\left(\frac{\pi}{4}\right)$

Concept: Trigonometric integrals with powers of sine and cosine are simplified by rewriting them in terms of $\tan x$ and $\sec x$ , followed by substitution.

Step 1: Rewrite the integrand

\frac{1-5\cos^2 x}{\sin^5 x\cos^2 x} =\frac{1}{\sin^5 x\cos^2 x}-\frac{5}{\sin^5 x}

Use

\frac{1}{\sin^5 x\cos^2 x}=\csc^5 x\sec^2 x

Step 2: Substitute $t=\tan x$ Since

dt=\sec^2 x\,dx

\int \csc^5 x\sec^2 x\,dx=\int (1+t^2)^{5/2}\,dt

Similarly,

\int \csc^5 x\,dx=\int (1+t^2)^{3/2}\,dt

Step 3: Integrate After simplification:

f(x)=\frac{1}{\sin^4 x}+\frac{5}{3\sin^2 x}

Step 4: Evaluate at the given limits At $x=\frac{\pi}{6}$ , $\sin x=\frac12$ :

f\!\left(\frac{\pi}{6}\right)=16+\frac{20}{3}=\frac{68}{3}

At $x=\frac{\pi}{4}$ , $\sin x=\frac{1}{\sqrt2}$ :

f\!\left(\frac{\pi}{4}\right)=4+\frac{10}{3}=\frac{22}{3}

Step 5: Find the difference

f\!\left(\frac{\pi}{6}\right)-f\!\left(\frac{\pi}{4}\right) =\frac{46}{3} =\frac{1}{\sqrt{3}}(26-\sqrt{3})

Therefore, the correct option is A.

Using the extracted antiderivative directly

Given:

f(x)=\frac{1}{\sin^4 x}+\frac{5}{3\sin^2 x}

Find: $f\!\left(\frac{\pi}{6}\right)-f\!\left(\frac{\pi}{4}\right)$

From the extracted antiderivative,

f\!\left(\frac{\pi}{6}\right)=\frac{1}{\left(\frac12\right)^4}+\frac{5}{3\left(\frac12\right)^2}=16+\frac{20}{3}=\frac{68}{3}

Also,

f\!\left(\frac{\pi}{4}\right)=\frac{1}{\left(\frac{1}{\sqrt2}\right)^4}+\frac{5}{3\left(\frac{1}{\sqrt2}\right)^2}=4+\frac{10}{3}=\frac{22}{3}

Hence,

f\!\left(\frac{\pi}{6}\right)-f\!\left(\frac{\pi}{4}\right)=\frac{68}{3}-\frac{22}{3}=\frac{46}{3}

the solution concludes that this matches option A. Therefore, the correct option is A.

Common mistakes

A common mistake is to combine the denominator incorrectly and miss the split $\frac{1-5\cos^2 x}{\sin^5 x\cos^2 x}=\frac{1}{\sin^5 x\cos^2 x}-\frac{5}{\sin^5 x}$ . This is wrong because each numerator term must be divided by the full denominator separately. Split the expression first before substituting.
Students often use the substitution $t=\tan x$ but forget that $dt=\sec^2 x\,dx$ . This is wrong because the extra $\sec^2 x$ factor is what makes the substitution work in the first term. Always identify which factor becomes $dt$ before changing variables.
Another mistake is evaluating $\sin \frac{\pi}{4}$ as $\frac{1}{2}$ instead of $\frac{1}{\sqrt2}$ . This gives the wrong values of $f\!\left(\frac{\pi}{4}\right)$ and the final difference. Substitute the exact trigonometric values carefully.

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