JEE Main 2026 Jan 28 Shift 1, Mathematics Q11: Integration by Parts

Split: $\int \frac{dx}{{\sin }^{5}x{\cos }^{2}x}-5\int \frac{dx}{{\sin }^{5}x}$.

Rewrite the first integral as $\int \frac{{\sec }^{2}x}{{\sin }^{5}x}dx$. Apply integration by parts with $u=\frac{1}{{\sin }^{5}x}$ and $dv={\sec }^{2}xdx$ (so $v=\tan x$):

$\int \frac{{\sec }^{2}x}{{\sin }^{5}x}dx=\frac{\tan x}{{\sin }^{5}x}-\int \tan x\cdot \left(-\frac{5\cos x}{{\sin }^{6}x}\right)dx=\frac{\tan x}{{\sin }^{5}x}+5\int \frac{dx}{{\sin }^{5}x}$.

Substituting back, the $\int dx/{\sin }^{5}x$ pieces cancel exactly:

$f(x)=\frac{\tan x}{{\sin }^{5}x}$.

Evaluate at the two points:

$f\left(\frac{\pi }{6}\right)=\frac{1/\sqrt{3}}{(1/2{)}^5}=\frac{32}{\sqrt{3}}$.

$f\left(\frac{\pi }{4}\right)=\frac{1}{(1/\sqrt{2}{)}^5}=2^{5/2}=4\sqrt{2}$.

Difference: $\frac{32}{\sqrt{3}}-4\sqrt{2}=\frac{32-4\sqrt{6}}{\sqrt{3}}=\frac{4}{\sqrt{3}}(8-\sqrt{6})$.

Option (2).