PrepRung
Sign inSign up free

JEE Main 2022 · Mathematics · Indefinite and Definite Integrals · hard

∫0^π(xsin x)/(1+cos ^2x) dx equals

Single correct option

∫0πxsin⁡x1+cos⁡2x dx\displaystyle\int_0^{\pi}\dfrac{x\sin x}{1+\cos^2x}\,dx equals
  1. A. π28\dfrac{\pi^2}{8}
  2. B. π24\dfrac{\pi^2}{4}
  3. C. π22\dfrac{\pi^2}{2}
  4. D. π4\dfrac{\pi}{4}
Show answer and solution

Answer: (B) π24\dfrac{\pi^2}{4}

Let I=∫0πxsin⁡x1+cos⁡2xdxI=\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}dx. Replace xx by π−x\pi-x: sin⁡(π−x)=sin⁡x\sin(\pi-x)=\sin x and cos⁡2(π−x)=cos⁡2x\cos^2(\pi-x)=\cos^2x, so I=∫0π(π−x)sin⁡x1+cos⁡2xdxI=\int_0^{\pi}\frac{(\pi-x)\sin x}{1+\cos^2x}dx. Adding, 2I=π∫0πsin⁡x1+cos⁡2xdx2I=\pi\int_0^{\pi}\frac{\sin x}{1+\cos^2x}dx. With u=cos⁡xu=\cos x the last integral is ∫−11du1+u2=π2\int_{-1}^{1}\frac{du}{1+u^2}=\frac\pi2. Therefore 2I=π222I=\frac{\pi^2}{2} and I=π24I=\frac{\pi^2}{4}. Numerical integration gives 2.46742.4674, confirming it.
Report a mistake

Original question written in the JEE Main 2022 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake