$\int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^{3}+\cos ^{2}(x+3 \pi)\right] d x$ is equal to

(B) $\frac{\pi^{4}}{32}+\frac{\pi}{2}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Definite Integrals

\int_{- 3 π /2}^{- π /2} [(x + π)^{3} + cos^{2} (x + 3 π)] d x

is equal to

Answer: (C)

Substitute

t = x + π

, so

d t = d x

, and the limits

x = - \frac{3 π}{2}, - \frac{π}{2}

become

t = - \frac{π}{2}, \frac{π}{2}

. Since

cos^{2} (x + 3 π) = cos^{2} (t + 2 π) = cos^{2} t

, the integral becomes

\int_{- π /2}^{π /2} [t^{3} + cos^{2} t] d t

. The term

t^{3}

is odd and integrates to zero over the symmetric interval, while

cos^{2} t

is even, so the value is

2 \int_{0}^{π /2} cos^{2} t d t = \int_{0}^{π /2} (1 + cos 2 t) d t = [t + \frac{s i n 2 t}{2}]_{0}^{π /2} = \frac{π}{2}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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