Lời giải:
Ta có: \(I=\int \sin (\cos ^4x+x)dx=\int \sin x\cos ^4xdx+\int x\sin xdx\)
Xét \(\int \sin x\cos ^4xdx=\int \cos ^4x(\sin xdx)=-\int \cos ^4xd(\cos x)\)
\(=-\frac{\cos ^5x}{5}+c\)
Xét \(\int x\sin xdx\)
Đặt \(\left\{\begin{matrix} u=x\\ dv=\sin xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\int \sin xdx=-\cos x\end{matrix}\right.\)
\(\Rightarrow \int x\sin xdx=-x\cos x+\int \cos xdx=-xcos x+\sin x+c\)
Do đó:
\(I=\frac{-\cos ^5x}{5}+\sin x-x\cos x+c\)
\(I=\int x.\sin xdx+\int\cos^4x.\sin xdx\)
= \(-\int xd\left(\cos x\right)-\int\cos^4xd\left(\cos x\right)\)
= \(-x.\cos x+\int\cos xdx-\dfrac{\cos^5x}{5}\)
= \(-x.\cos x+\sin x-\dfrac{\cos^5x}{5}+C\)
