7.1 Integration By Parts/45

${\begin{aligned}&\int _{0}^{\frac {\pi }{2}}sin^{n}(x)dx={\frac {n-1}{n}}\int _{0}^{\frac {\pi }{2}}sin^{n-2}(x)dx\\[2ex]&=-{\frac {1}{n}}cos(x)sin^{n-1}(x)+{\frac {n-1}{n}}{\bigg |}_{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}sin^{n-2}(x)dx\\[2ex]&=-{\frac {1}{n}}cos({\frac {\pi }{2}})sin^{n-1}({\frac {\pi }{2}})+{\frac {n-1}{n}}\int _{0}^{\frac {\pi }{2}}sin^{n-2}(x)dx\\[2ex]&=-{\frac {1}{n}}(0)(1)+{\frac {n-1}{n}}\int _{0}^{\frac {\pi }{2}}sin^{n-2}(x)dx\\[2ex]&={\frac {n-1}{n}}\int _{0}^{\frac {\pi }{2}}sin^{n-2}(x)dx\\[2ex]\end{aligned}}$

${\begin{aligned}&\int _{0}^{\frac {\pi }{2}}sin^{3}(x)dx\\[2ex]&={\frac {3-1}{3}}\int _{0}^{\frac {\pi }{2}}sin(x)dx\\[2ex]&={\frac {2}{3}}\int _{0}^{\frac {\pi }{2}}sin(x)dx\\[2ex]&={\frac {2}{3}}[-cos(x){\bigg |}_{0}^{\frac {\pi }{2}}\\[2ex]&={\frac {2}{3}}[-cos({\frac {\pi }{2}})-(-cos(0))]\\[2ex]&={\frac {2}{3}}[-1(0)-(-1)]\\[2ex]&={\frac {2}{3}}[1]\\[2ex]&={\frac {2}{3}}\\[2ex]\end{aligned}}$

${\begin{aligned}&\int _{0}^{\frac {\pi }{2}}sin^{2n+1}(x)dx\\[2ex]&={\frac {2n+1-1}{2n+1}}\int _{0}^{\frac {\pi }{2}}sin^{2n+1-2}(x)dx\\[2ex]&={\frac {2n}{2n+1}}\int _{0}^{\frac {\pi }{2}}sin^{2n-1}(x)dx\\[2ex]\end{aligned}}$

${\begin{aligned}\int _{0}^{\frac {\pi }{2}}sin^{5}(x)dx&={\frac {5-1}{5}}\int _{0}^{\frac {\pi }{2}}sin^{3}(x)dx\\[2ex]&={\frac {4}{5}}\int _{0}^{\frac {\pi }{2}}(1-cos^{2}(x))sin(x)dx\\[2ex]&=-\int _{1}^{0}(1-u^{2})\cdot du\\[2ex]&={\frac {4}{5}}\left[u-{\frac {u^{3}}{3}}\right]{\bigg |}_{0}^{1}\\[2ex]&u=cos(x)\\[2ex]&du=-sin(x)\\[2ex]&dx={\frac {du}{-sin(x)}}\\[2ex]&\int sin^{3}(x)dx\\[2ex]&={\frac {4}{5}}[1-{\frac {1}{3}}]\\[2ex]&={\frac {8}{15}}\end{aligned}}$