MHTCET 16 Apr 2026 Shift 2 Paper

Concept:The given expression is an integral of the form $e^{\cos x}$ times a function, solved by checking which option differentiates back to the integrand.Explanation:Given $f(x)=\cos x$, the integral becomes$$\int \frac{e^{\cos x}\left(x \sin^3 x+\cos x\right)}{1-\cos^2 x}\,dx$$Using $1-\cos^2 x=\sin^2 x$, we get$$\int \frac{e^{\cos x}\left(x \sin^3 x+\cos x\right)}{\sin^2 x}\,dx$$Dividing each term in the numerator by $\sin^2 x$,$$\int e^{\cos x}\left(x\sin x+\frac{\cos x}{\sin^2 x}\right)dx$$Now differentiate option A: $e^{\cos x}\left(-\operatorname{cosec}x-x\right)$.Using the product rule and $\frac{d}{dx}(\operatorname{cosec}x)=-\operatorname{cosec}x\cot x$,$$\frac{d}{dx}\left[e^{\cos x}(-\operatorname{cosec}x-x)\right]$$$$=e^{\cos x}(-\sin x)(-\operatorname{cosec}x-x)+e^{\cos x}(\operatorname{cosec}x\cot x-1)$$$$=e^{\cos x}\left(1+x\sin x+\frac{\cos x}{\sin^2 x}-1\right)$$$$=e^{\cos x}\left(x\sin x+\frac{\cos x}{\sin^2 x}\right)$$This exactly matches the integrand, so the antiderivative is correct.Answer:$$\int \frac{e^{f(x)}\left(x \sin^3 x+f(x)\right)}{1-(f(x))^2}\,dx=e^{f(x)}(-\operatorname{cosec}x-x)+c$$Hence, the correct option is $A$.