A) \[\left( \frac{{{\pi }^{4}}}{32} \right)+\left( \frac{\pi }{2} \right)\]
B) \[\frac{\pi }{2}\]
C) \[\left( \frac{\pi }{4} \right)-1\]
D) \[\frac{{{\pi }^{4}}}{32}\]
Correct Answer: B
Solution :
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 \[2l=\int\limits_{-3\pi /2}^{-\pi /2}{2{{\cos }^{2}}x}dx\] \[=\int\limits_{-3\pi /2}^{-\pi /2}{(1+\cos 2x)}dx\] \[=\left[ x+\frac{\sin 2x}{2} \right]_{-3\pi /2}^{-\pi /2}\] \[=\left[ -\frac{\pi }{2}+\frac{\sin (-\pi )}{2}-\left( -\frac{3\pi }{2}+\frac{\sin (-3\pi )}{2} \right) \right]\] \[=-\frac{\pi }{2}+\frac{3\pi }{2}=\pi \] \[\Rightarrow \,\,\,l=\frac{\pi }{2}\]
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