Question

hi, I want to know how to integrate [pi(cosx)^4+x(cosx)^4 ]

Answer 1

well you can devide the integral into 2 parts , the first one is (pi cos^4(x))
\[\int\limits_{}^{} \pi \cos ^{4}(x) =\pi \int\limits_{}^{} \cos ^{2}(x) * (1-\sin ^{2}(x))\]
\[\pi \int\limits_{}^{} \cos ^{2}(x)-\cos ^{2}(x) \sin ^{2}(x) = \frac{ \pi }{ 2 } \int\limits_{}^{} 1+\cos(2x) - \frac{ 1 }{ 4 }(1-\cos(2x))\]
I just used the identities to get there,,,
the other one is \[\int\limits_{}^{} xcos ^{4}(x)\] it has the same idea just you can use integration by parts to eliminate x

Answer 2

\[\int{}^{}\pi \space cos^{4}x\space dx =\pi\int{}^{}cos^{2}x(1-sin^{2}x) \space dx\\= \pi \int\limits_{}^{} \cos ^{2}(x)-\cos ^{2}(x) \sin ^{2}(x)\space dx\\=\pi\int{}{}cos^{2}(x)dx-\pi\int{}^{}\frac{1}{4}(1+cos(2x))(1-cos\space(2x))\space dx\\=\pi\int{}{}\frac{1+cos(2x)}{2}dx-\frac{\pi}{4}\int{}^{}1-cos^{2}(2x)\space dx\\=\pi\int{}{}\frac{1+cos(2x)}{2}dx-\frac{1}{4}\int{}^{}sin^{2}(2x)\space dx\\=\pi\int{}{}\frac{1+cos(2x)}{2}dx-\frac{\pi}{4}\int{}{}\frac{1-cos(4x)}{2}dx\\=\frac{\pi}{2}\left( x+\frac{sin(2x)}{2} \right) -\frac{\pi}{8}\left(x-\frac{sin(4x)}{4} \right)\\=\frac{\pi}{32} \left( \space 12x+8sin(2x)+sin(4x)\space \right) \]