`int_0^(pi/2) sin^7(theta) cos^5(theta) d theta` Evaluate the integral

Thursday, September 28, 2017

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\sin}}^{{7}}{\left(\theta\right)}{{\cos}}^{{5}}{\left(\theta\right)}{d}\theta$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\sin}}^{{7}}{\left(\theta\right)}{{\cos}}^{{5}}{\left(\theta\right)}{d}\theta$

Let's first compute the indefinite integral by rewriting the integrand,

$\displaystyle\int{{\sin}}^{{7}}{\left(\theta\right)}{{\cos}}^{{5}}{\left(\theta\right)}{d}\theta=\int{{\sin}}^{{6}}{\left(\theta\right)}{\sin{{\left(\theta\right)}}}{{\cos}}^{{5}}{\left(\theta\right)}{d}\theta$

Now use the identity: $\displaystyle{{\sin}}^{{2}}{\left({x}\right)}={1}-{{\cos}}^{{2}}{\left({x}\right)}$

$\displaystyle=\int{{\left({1}-{{\cos}}^{{2}}{\left(\theta\right)}\right)}}^{{3}}{\sin{{\left(\theta\right)}}}{{\cos}}^{{5}}{\left(\theta\right)}{d}\theta$

Now apply integral substitution,

Let $\displaystyle{u}={\cos{{\left(\theta\right)}}}$

$\displaystyle\Rightarrow{d}{u}=-{\sin{{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\int{{\left({1}-{{u}}^{{2}}\right)}}^{{3}}{{u}}^{{5}}{\left(-{d}{u}\right)}$

$\displaystyle=-\int{{\left({1}-{{u}}^{{2}}\right)}}^{{3}}{{u}}^{{5}}{d}{u}$

$\displaystyle=-\int{\left({1}-{{u}}^{{6}}-{3}{{u}}^{{2}}+{3}{{u}}^{{4}}\right)}{{u}}^{{5}}{d}{u}$

$\displaystyle=-\int{\left({{u}}^{{5}}-{{u}}^{{11}}-{3}{{u}}^{{7}}+{3}{{u}}^{{9}}\right)}{d}{u}$

$\displaystyle=-\int{{u}}^{{5}}{d}{u}+\int{{u}}^{{11}}{d}{u}+{3}\int{{u}}^{{7}}{d}{u}-{3}\int{{u}}^{{9}}{d}{u}$

$\displaystyle=-\frac{{{u}}^{{6}}}{{6}}+\frac{{{u}}^{{12}}}{{12}}+{3}{\left(\frac{{{u}}^{{8}}}{{8}}\right)}-{3}{\left(\frac{{{u}}^{{10}}}{{10}}\right)}$

$\displaystyle=\frac{{{u}}^{{12}}}{{12}}-\frac{{3}}{{10}}{{u}}^{{10}}+\frac{{3}}{{8}}{{u}}^{{8}}-\frac{{1}}{{6}}{{u}}^{{6}}$

Substitute back $\displaystyle{u}={\cos{{\left({x}\right)}}}$

$\displaystyle=\frac{{1}}{{12}}{{\cos}}^{{12}}{\left({x}\right)}-\frac{{3}}{{10}}{{\cos}}^{{10}}{\left({x}\right)}+\frac{{3}}{{8}}{{\cos}}^{{8}}{\left({x}\right)}-\frac{{1}}{{6}}{{\cos}}^{{6}}{\left({x}\right)}$

Add a constant C to the solution,

Now let's evaluate the definite integral,

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\sin}}^{{7}}{\left(\theta\right)}{{\cos}}^{{5}}{\left(\theta\right)}{d}\theta={{\left[\frac{{1}}{{12}}{{\cos}}^{{12}}{\left({x}\right)}-\frac{{3}}{{10}}{{\cos}}^{{10}}{\left({x}\right)}+\frac{{3}}{{8}}{{\cos}}^{{8}}{\left({x}\right)}-\frac{{1}}{{6}}{{\cos}}^{{6}}{\left({x}\right)}\right]}_{{0}}^{{\frac{\pi}{{2}}}}}$

$\displaystyle={\left[\frac{{1}}{{12}}{{\cos}}^{{12}}{\left(\frac{\pi}{{2}}\right)}-\frac{{3}}{{10}}{{\cos}}^{{10}}{\left(\frac{\pi}{{2}}\right)}+\frac{{3}}{{8}}{{\cos}}^{{8}}{\left(\frac{\pi}{{2}}\right)}-\frac{{1}}{{6}}{{\cos}}^{{6}}{\left({x}\right)}\right]}-{\left[\frac{{1}}{{12}}{{\cos}}^{{12}}{\left({0}\right)}-\frac{{3}}{{10}}{{\cos}}^{{10}}{\left({0}\right)}+\frac{{3}}{{8}}{{\cos}}^{{8}}{\left({0}\right)}-\frac{{1}}{{6}}{{\cos}}^{{6}}{\left({0}\right)}\right]}$

Plug in the values of $\displaystyle{\cos{{\left(\frac{\pi}{{2}}\right)}}}={0}.{\cos{{\left({0}\right)}}}={1}$

$\displaystyle={\left[{0}\right]}-{\left[\frac{{1}}{{12}}-\frac{{3}}{{10}}+\frac{{3}}{{8}}-\frac{{1}}{{6}}\right]}$

$\displaystyle=-{\left[\frac{{{10}-{36}+{45}-{20}}}{{120}}\right]}$

$\displaystyle=-{\left[-\frac{{1}}{{120}}\right]}$

$\displaystyle=\frac{{1}}{{120}}$