∫ 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}}}
New upper limit: π = ( π ) 2 {\displaystyle \pi =({\sqrt {\pi }})^{2}} New lower limit: 0 = ( 0 ) 2 {\displaystyle 0=(0)^{2}}
∫ 0 π x cos ( x 2 ) d x = ∫ 0 π ( x d x ) cos ( x 2 ) = ∫ 0 π ( 1 2 d u ) cos ( u ) = 1 2 ∫ 0 π cos ( u ) d u = 1 2 sin ( u ) | 0 π = 1 2 sin ( π ) − 1 2 sin ( 0 ) = 0 {\displaystyle {\begin{aligned}\int _{0}^{\sqrt {\pi }}x\cos {(x^{2})}\,dx&=\int _{0}^{\sqrt {\pi }}(xdx)\cos {(x^{2})}\\[2ex]&=\int _{0}^{\pi }\left({\frac {1}{2}}du\right)\cos {(u)}={\frac {1}{2}}\int _{0}^{\pi }\cos {(u)}du\\[2ex]&={\frac {1}{2}}\sin {(u)}{\bigg |}_{0}^{\pi }\\[2ex]&={\frac {1}{2}}\sin {(\pi )}-{\frac {1}{2}}\sin {(0)}\\[2ex]&=0\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})}\\[2ex]&=\int _{0}^{\pi }\left({\frac {1}{2}}du\right)\cos {(u)}={\frac {1}{2}}\int _{0}^{\pi }\cos {(u)}du\\[2ex]&={\frac {1}{2}}\sin {(u)}{\bigg |}_{0}^{\pi }\\[2ex]&={\frac {1}{2}}\sin {(\pi )}-{\frac {1}{2}}\sin {(0)}\\[2ex]&=0\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/d30fc1f61658d5bb276f432ee3bb7718a814f5c9)