MHTCET 16 Apr 2026 Shift 2 Paper

Concept:Use the symmetry of odd and even functions to simplify the integral, then apply the standard property $\int_{0}^{\pi}xf(x)\,dx=\frac{\pi}{2}\int_{0}^{\pi}f(x)\,dx$ when $f(\pi-x)=f(x)$.Explanation:Let$$I=\int_{-\pi}^{\pi}\frac{2x(1+\sin x)}{1+\cos^2 x}\,dx.$$Split the integrand:$$I=\int_{-\pi}^{\pi}\frac{2x}{1+\cos^2 x}\,dx+\int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^2 x}\,dx.$$Since $1+\cos^2 x$ is even and $2x$ is odd, the first integrand is odd. Therefore, its integral over $[-\pi,\pi]$ is $0$.Thus,$$I=\int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^2 x}\,dx.$$The integrand $\frac{2x\sin x}{1+\cos^2 x}$ is even, so$$I=2\int_{0}^{\pi}\frac{2x\sin x}{1+\cos^2 x}\,dx=4\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2 x}\,dx.$$Let$$J=\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2 x}\,dx.$$Here, $f(x)=\frac{\sin x}{1+\cos^2 x}$ satisfies $f(\pi-x)=f(x)$. Hence,$$J=\frac{\pi}{2}\int_{0}^{\pi}\frac{\sin x}{1+\cos^2 x}\,dx.$$Now substitute $u=\cos x$, so $du=-\sin x\,dx$.Limits change from $x=0\to u=1$ and $x=\pi\to u=-1$.Thus,$$\int_{0}^{\pi}\frac{\sin x}{1+\cos^2 x}\,dx=\int_{-1}^{1}\frac{du}{1+u^2}=\left[\tan^{-1}u\right]_{-1}^{1}=\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)=\frac{\pi}{2}.$$Therefore,$$J=\frac{\pi}{2}\cdot\frac{\pi}{2}=\frac{\pi^2}{4}.$$Finally,$$I=4J=4\cdot\frac{\pi^2}{4}=\pi^2.$$Answer:$$\pi^2$$So the correct option is Option B.