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

Question

Aicharya · Accepted Answer

$ I=\int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^3+\cos ^2(x+3 \pi)ight] d x $ Put $\mathrm{x}+\pi=\mathrm{t}$ $ \begin{aligned} & I=\int_{-\pi / 2}^{\pi / 2}\left[t^3+\cos ^2 tight] d t=2 \int_0^{\pi / 2} \cos ^2 t d t \ & =\int_0^{\pi / 2}(1+\cos 2 t) d t=\frac{\pi}{2}+0 . \end{aligned} $

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

Solution