_-3/2^-/2 [(x+)^3+^2(x+3)],dx is equal to:

Question

$\displaystyle\int_{-3\pi/2}^{-\pi/2} [(x+\pi)^3+\cos^2(x+3\pi)]\,dx$ is equal to:

Accepted Answer

$$I = \int\limits_{ - {{3\pi } \over 2}}^{ - {\pi  \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} \,dx$$

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 $$\left. {} \right]$$

$$ = \int\limits_0^{{\pi  \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi  \over 2} + 0$$

Solutions · 1