MHTCET 25 Apr 2025 Shift 2 Paper

Concept:The integrand $(\sin x)^{-4} = \csc^4 x$ is an even function, but it has an infinite discontinuity at $x = 0$ inside the interval, so the integral is improper and must be checked for convergence.Explanation:By evenness, $I = 2\int_0^{\pi/4}\csc^4 x\,dx$.Rewrite $\csc^4 x = \frac{(1+\tan^2 x)\sec^2 x}{\tan^4 x}$ and put $t = \tan x$, so $\sec^2 x\,dx = dt$.When $x = 0$, $t = 0$; when $x = \frac{\pi}{4}$, $t = 1$.Hence $I = 2\int_0^1 \frac{1+t^2}{t^4}\,dt = 2\int_0^1 (t^{-4} + t^{-2})\,dt$.As $t \to 0^+$, both $t^{-4}$ and $t^{-2}$ tend to infinity, so both improper integrals $\int_0^1 t^{-4}\,dt$ and $\int_0^1 t^{-2}\,dt$ diverge to $+\infty$.The mistake in the original solution was substituting $t = 0$ into $t^{-3}$ and $t^{-1}$ as though they were zero; actually, they are undefined and infinite at $t = 0$.Also, since the integrand is non-negative on $(-\pi/4, \pi/4)$, the integral cannot equal a negative number like $-\frac{8}{3}$.Answer:The given improper integral diverges to $+\infty$; hence it has no finite value, so none of the options A–D is correct.