MHTCET 10 May 2024 Shift 2 Paper

Concept:The integrand $(\sin x)^{-4}$ becomes infinite at $x=0$, which lies inside the integration interval. Hence this must be handled as an improper integral.Explanation:Let $I = \int_{-\pi/4}^{\pi/4} (\sin x)^{-4}\, dx$.Since $(\sin x)^{-4}$ is even, we may write $I = 2\int_{0}^{\pi/4} (\sin x)^{-4}\, dx$.Near $x=0$, we have $\sin x \sim x$, so $(\sin x)^{-4} \sim x^{-4}$.But $\int_{0}^{\epsilon} x^{-4}\, dx$ diverges to $+\infty$, so the given integral also diverges.Using the substitution $t = \tan x$, the integral becomes $I = 2\int_{0}^{1} \frac{1+t^2}{t^4}\, dt$.The term $\int_{0}^{1} t^{-4}\, dt$ is not finite, so the value of $I$ is not finite either.Thus the integral does not converge to any real number.Answer:The integral diverges to $+\infty$. Therefore, none of the given finite options is correct.