• # question_answer $\int_{-3\pi /2}^{-\pi /2}{[{{(x+\pi )}^{3}}+{{\cos }^{2}}(x+3\pi )]}dx$is equal to     AIEEE  Solved  Paper-2006 A) $\left( \frac{{{\pi }^{4}}}{32} \right)+\left( \frac{\pi }{2} \right)$                               B) $\frac{\pi }{2}$                               C)        $\left( \frac{\pi }{4} \right)-1$  D)        $\frac{{{\pi }^{4}}}{32}$

Let $l=\int\limits_{-3\pi /2}^{-\pi /2}{[{{(x+\pi )}^{3}}+{{\cos }^{2}}(x+3\pi )]}dx$ Or           $l=\int\limits_{-3\pi /2}^{-\pi /2}{{{(x+\pi )}^{3}}+{{\cos }^{2}}x]}dx$ and $l=\int\limits_{-3\pi /2}^{-\pi /2}{\left[ {{\left( -\frac{\pi }{2}-\frac{3\pi }{2}-x+\pi \right)}^{3}} \right.}$ $\left. +{{\cos }^{2}}\left( -\frac{\pi }{2}-\frac{3\pi }{2}-x \right) \right]dx$ $\Rightarrow$$l=\int\limits_{-3\pi /2}^{-\pi /2}{[-{{(x+\pi )}^{3}}+{{\cos }^{2}}(-2\pi -x)}]dx$ $l=\int\limits_{-3\pi /2}^{-\pi /2}{[-{{(x+\pi )}^{3}}+{{\cos }^{2}}x}]dx$ ?(ii) On adding Eqs. (i) and (ii), we get $2l=\int\limits_{-3\pi /2}^{-\pi /2}{2{{\cos }^{2}}x}dx$ $=\int\limits_{-3\pi /2}^{-\pi /2}{(1+\cos 2x)}dx$ $=\left[ x+\frac{\sin 2x}{2} \right]_{-3\pi /2}^{-\pi /2}$ $=\left[ -\frac{\pi }{2}+\frac{\sin (-\pi )}{2}-\left( -\frac{3\pi }{2}+\frac{\sin (-3\pi )}{2} \right) \right]$ $=-\frac{\pi }{2}+\frac{3\pi }{2}=\pi$ $\Rightarrow \,\,\,l=\frac{\pi }{2}$