Question

I = ${\int }_{-\frac{-3\pi }{2}}^{\frac{\pi }{2}}$ $\left[\right(x+\pi {)}^3+co{s}^3(x+3\pi )]$ dx

Answer 1

$I={\int }_{\frac{-3\pi }{2}}^{+\pi /2}\left[\right(x+\pi {)}^3+{\cos }^3(x+3\pi )]dx$

$I={\int }_{\frac{-3\pi }{2}}^{+\pi /2}(x+\pi {)}^{3}dx+{\int }_{\frac{-3\pi }{2}}^{+\pi /2}{\cos }^3(x+3\pi )dx$

$I=\frac{(x+\pi {)}^4}{4}{|}_{\frac{-3\pi }{2}}^{\pi /2}+{\int }_{\frac{-3\pi }{2}}^{+n/2}{\cos }^3(x+3\pi )dx$

$I=\frac{1}{4}({\left(\frac{3\pi }{2}\right)}^4-{\left(\frac{-\pi }{2}\right)}^4)+I_1$

$I=\frac{1}{4}\left[n^4\right][{\left(\frac{3}{2}\right)}^4-\frac{1}{2^4})+I_1$

$I=\frac{n^4}{2^6}[3^4-1]+I_1$

where $I_1={\int }_{\frac{-3\pi }{2}}^{+\pi /2}{\cos }^3(x+3\pi )dx$

$t=x+3\pi$

$I_1={\int }_{\frac{3\pi }{2}}^{+\pi /2}{\cos }^3+dt={\int }_{\frac{3\pi }{2}}^{+\pi /2}\left(\frac{\cos 3t+3\cos t}{4}\right)dt$

$I_1={\int }_{\frac{3\pi }{2}}^{+\pi /2}\frac{\cos 3t}{4}\frac{3}{4}{\int }_{\frac{3\pi }{2}}^{+\pi /2}\cos tdt=\frac{1}{4}\times \left(\frac{1}{3}\right)\sin 3t{|}_{\frac{3\pi }{2}}^{+\pi /2}$ $+\frac{3}{4}\times \sin t{|}_{3\pi /2}^{+\pi /2}$

$I_1=\frac{1}{12}[\sin \left[\frac{21\pi }{2}\right]-\sin \left[\frac{9\pi }{2}\right]]+\frac{3}{4}[\sin \left(\frac{7\pi }{2}\right)-\sin \left(\frac{3\pi }{2}\right)]$

$I_1=\frac{1}{12}[1-1]+\frac{3}{4}\left[\right(-1)-(-1\left)\right]=0$

$I=\frac{{\pi }^4}{2^6}[3^4-1]$