Solveeit Logo
Login

Question

Mathematics Question on integral

∫−3π/2−π/2\int\limits^{-\pi/2}_{{-3\pi/2}} [(x+π)3+cos2(x+3π)]dx\left[\left(x+\pi\right)^{3}+cos^{2} \left(x+3\pi\right)\right]dx is equal to :

A

(π432)+(π2)\left(\frac{\pi^{4}}{32}\right)+\left(\frac{\pi}{2}\right)

B

π2\frac{\pi}{2}

C

(π4)−1\left(\frac{\pi}{4}\right)-1

D

π432\frac{\pi^{4}}{32}

Answer

π2\frac{\pi}{2}

Explanation

Solution

Let ∫−3π/2−π/2\int\limits^{-\pi/2}_{{-3\pi/2}} [(x+π)3+cos2(x+3π)]dx...(i)\left[\left(x+\pi\right)^{3}+cos^{2} \left(x+3\pi\right)\right]dx ...(i) and I=∫−3π/2−π/2I=\int\limits^{-\pi/2}_{{-3\pi/2}} [(−π2−3π2−x+π)3+cos2(−π2−3π2−x+3π)]dx\left[\left(-\frac{\pi}{2}-\frac{3\pi}{2}-x+\pi\right)^{3}+cos^{2}\left(-\frac{\pi}{2}-\frac{3\pi}{2}-x+3\pi\right)\right]dx ⇒\Rightarrow I=∫−3π/2−π/2I=\int\limits^{-\pi/2}_{{-3\pi/2}} [−(x+π)3+cos2(π−x)]dx...(ii)\left[-\left(x+\pi\right)^{3}+cos^{2}\left(\pi-x\right)\right]dx ...\left(ii\right) On adding Eqs. (i) and (ii), we get 2I=∫−3π/2−π/22I=\int\limits^{-\pi/2}_{{-3\pi/2}} 2 cos2 x dx2\,cos^{2}\,x\,dx =∫−3π/2−π/2=\int\limits^{-\pi/2}_{{-3\pi/2}} (1+cos 2x)dx\left(1+cos\,2x\right)dx =[x+sin 2x2]−3π/2−π/2=\left[x+\frac{sin\,2x}{2}\right]^{-\pi/2}_{_{_{-3\pi/2}}} =−π2+3π2=π=-\frac{\pi}{2}+\frac{3\pi}{2}=\pi ⇒I=π2\Rightarrow I=\frac{\pi}{2}