∫ 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 7 4 + 3 x d x = ∫ 4 25 u d u = ∫ ( d u ) sin ( u ) = ∫ sin ( u ) d u = − cos ( u ) + C = − cos ( ln ( x ) ) + C {\displaystyle {\begin{aligned}\int _{0}^{7}{\sqrt {4+3x}}\,dx=\int _{4}^{25}{\sqrt {u}}\,du\\[2ex]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\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}^{7}{\sqrt {4+3x}}\,dx=\int _{4}^{25}{\sqrt {u}}\,du\\[2ex]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/af493f4689d83b65aef378316215058b6ecde692)