7.1 Integration By Parts/45

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

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

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

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

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\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] &= -\int 1-u^2 du \\[2ex] &= \frac{4}{5} [-cos(x) + \frac{1}{3} cos^3(x)] \\[2ex] &= \frac{4}{5} [-cos(\frac{\pi}{2}) + \frac{1}{3} cos^3 (\frac{\pi}{2}) - (-cos(0) + \frac{1}{3} cos^3 (0)) \\[2ex] &= \frac{4}{5} [(-1)(0) + \frac{1}{3} (0)^3 - ((-1)(1) +\frac{1}{3} cos^3(1) \\[2ex]

&= \frac{4}{5} [-(-1+ \frac{1}{3}(1)] \\[2ex] &= \frac{4}{5}[\frac{2}{3}] \\[2ex] &= \frac{8}{15} \\[2ex]

\end{align}