7.1 Integration By Parts/45

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^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{align} }

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

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

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