∫ 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 ) = ∫ 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 ) ) {\displaystyle {\begin{aligned}\int _{0}^{\sqrt {\pi }}x\cos {(x^{2})}\,dx&=\int _{0}^{\sqrt {\pi }}(xdx)\cos {(x^{2})}\\[2ex]&=\int _{0}^{\pi }({\frac {1}{2}}du)\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]\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 }({\frac {1}{2}}du)\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]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/eb91621dc612cdf30e632cc83aaff06a66393b96)