∫ 0 π x cos ( x 2 ) d x {\displaystyle \int _{0}^{\sqrt {\pi }}x\cos {(x^{2})}\,dx}
u = x 2 d u = 2 x d x 1 2 d u = x d x {\displaystyle {\begin{aligned}u&=x^{2}\\[2ex]du&=2xdx\\[2ex]{\frac {1}{2}}du&=xdx\\[2ex]\end{aligned}}}
∫ 0 π x cos ( x 2 ) d x = ∫ 0 π ( x d x ) cos ( x 2 ) = = ∫ ( 1 2 d u ) cos ( u ) = 1 2 ∫ cos ( u ) d u = 1 2 sin ( u ) + C = 1 2 sin ( x 2 ) + C {\displaystyle {\begin{aligned}\int _{0}^{\sqrt {\pi }}x\cos {(x^{2})}\,dx&=\int _{0}^{\sqrt {\pi }}(xdx)\cos {(x^{2})}=&=\int ({\frac {1}{2}}du)\cos {(u)}={\frac {1}{2}}\int \cos {(u)}du\\[2ex]&={\frac {1}{2}}\sin {(u)}+C\\[2ex]&={\frac {1}{2}}\sin {(x^{2})}+C\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=x^{2}\\[2ex]du&=2xdx\\[2ex]{\frac {1}{2}}du&=xdx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/18c922fd60a28dedb46a2819a6f7ab99b233dc17)
![{\displaystyle {\begin{aligned}\int _{0}^{\sqrt {\pi }}x\cos {(x^{2})}\,dx&=\int _{0}^{\sqrt {\pi }}(xdx)\cos {(x^{2})}=&=\int ({\frac {1}{2}}du)\cos {(u)}={\frac {1}{2}}\int \cos {(u)}du\\[2ex]&={\frac {1}{2}}\sin {(u)}+C\\[2ex]&={\frac {1}{2}}\sin {(x^{2})}+C\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a88c33a997701c63ffc9ca8b0f02658df5e06939)