JEE Main 2026 — 21 January, Morning Shift — Question 12

The integrand is even because $\pi$ is constant and $4x^{11}$ is odd, but $1-\sin(|x|+\pi/6)$ is even. So the integral equals $2\int_{0}^{\pi/6} \frac{\pi}{1-\sin(x+\pi/6)} \, dx.$ Let $t = x + \pi/6$, then $dt = dx$, limits: $x=0 \to t=\pi/6$, $x=\pi/6 \to t=\pi/3$. The integral becomes $2\pi \int_{\pi/6}^{\pi/3} \frac{dt}{1-\sin t}.$ Multiply numerator and denominator by $1+\sin t$: $\frac{1}{1-\sin t} = \frac{1+\sin t}{1-\sin^2 t} = \frac{1+\sin t}{\cos^2 t} = \sec^2 t + \sec t \tan t.$ Thus the integral is $2\pi \int_{\pi/6}^{\pi/3} (\sec^2 t + \sec t \tan t) \, dt = 2\pi \left[ \tan t + \sec t \right]_{\pi/6}^{\pi/3}.$ Evaluate: $\tan(\pi/3)=\sqrt{3}$, $\tan(\pi/6)=1/\sqrt{3}$, $\sec(\pi/3)=2$, $\sec(\pi/6)=2/\sqrt{3}$.

So $\left(\sqrt{3} + 2\right) - \left(\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}\right) = \sqrt{3} + 2 - \frac{3}{\sqrt{3}} = \sqrt{3} + 2 - \sqrt{3} = 2.$ Multiply by $2\pi$: $2\pi \times 2 = 4\pi.$