∫ x cos 5 x d x u = x d v = cos 5 x d x d u = d x v = 1 5 sin 5 x ∫ x cos 5 x d x = x ⋅ 1 5 sin 5 x − ∫ 1 5 sin 5 x = ∫ 1 5 sin 5 x ⋅ d x = 1 5 ∫ sin 5 x d x = 1 5 ⋅ 1 5 ( − cos 5 x ) = − 1 25 cos 5 x + C = 1 5 x sin 5 x + 1 25 cos 5 x + C {\displaystyle {\begin{aligned}\int \,x\cos 5xdx\\[2ex]u=x\qquad dv=\cos 5xdx\\[2ex]du=dx\qquad v={\frac {1}{5}}\sin 5x\\[2ex]\int \,x\cos 5xdx&=x\cdot {\frac {1}{5}}\sin 5x-\int {}^{}{\frac {1}{5}}\sin 5x\\[2ex]&=\int \,{\frac {1}{5}}\sin 5x\cdot \,dx\\[2ex]&={\frac {1}{5}}\int \,\sin 5xdx\\[2ex]&={\frac {1}{5}}\cdot {\frac {1}{5}}(-\cos 5x)\\[2ex]&=-{\frac {1}{25}}\cos 5x+C\\[2ex]&={\frac {1}{5}}x\sin 5x+{\frac {1}{25}}\cos 5x+C\end{aligned}}}
![{\displaystyle {\begin{aligned}\int \,x\cos 5xdx\\[2ex]u=x\qquad dv=\cos 5xdx\\[2ex]du=dx\qquad v={\frac {1}{5}}\sin 5x\\[2ex]\int \,x\cos 5xdx&=x\cdot {\frac {1}{5}}\sin 5x-\int {}^{}{\frac {1}{5}}\sin 5x\\[2ex]&=\int \,{\frac {1}{5}}\sin 5x\cdot \,dx\\[2ex]&={\frac {1}{5}}\int \,\sin 5xdx\\[2ex]&={\frac {1}{5}}\cdot {\frac {1}{5}}(-\cos 5x)\\[2ex]&=-{\frac {1}{25}}\cos 5x+C\\[2ex]&={\frac {1}{5}}x\sin 5x+{\frac {1}{25}}\cos 5x+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/0a58004e098cbef469bc59d255a8633b623b1e54)