JEE MAIN - Mathematics (2006 - No. 10)
$$\int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} dx$$ is equal to
$${{{\pi ^4}} \over {32}}$$
$${{{\pi ^4}} \over {32}} + {\pi \over 2}$$
$${\pi \over 2}$$
$${\pi \over 4} - 1$$
Explanation
$$I = \int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} \,dx$$
Put $$x + \pi = t$$
$$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{t^3} + {{\cos }^2}t} \right)dt} $$
$$ = 2\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}} tdt$$
$$\left[ {} \right.$$ using the property of even and odd function $$\left. {} \right]$$
$$ = \int\limits_0^{{\pi \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi \over 2} + 0$$
Put $$x + \pi = t$$
$$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{t^3} + {{\cos }^2}t} \right)dt} $$
$$ = 2\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}} tdt$$
$$\left[ {} \right.$$ using the property of even and odd function $$\left. {} \right]$$
$$ = \int\limits_0^{{\pi \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi \over 2} + 0$$
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