∫ 0 π x cos ( x 2 ) d x {\displaystyle \int _{0}^{\sqrt {\pi }}x\cos {(x^{2})}\,dx}
u = 4 + 3 x d u = 3 d x 1 3 d u = d x {\displaystyle {\begin{aligned}u&=4+3x\\[2ex]du&=3\,dx\\[2ex]{\frac {1}{3}}du&=dx\\[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&=4+3x\\[2ex]du&=3\,dx\\[2ex]{\frac {1}{3}}du&=dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/75fbdf9385d28dbfa67df6bec234667b6be64498)
![{\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)