7.1 Integration By Parts/45

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

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} & \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{align} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \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{align} }