# Evaluate the following integrals. \int_0^{\frac{\pi}{2}}\sin^7xdx

Evaluate the following integrals.
${\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{7}xdx$
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Ezra Herbert
Given ${\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{7}xdx$
${\mathrm{sin}}^{7}\left(x\right)={\mathrm{sin}}^{6}\left(x\right)\mathrm{sin}\left(x\right)$
and ${\mathrm{sin}}^{6}\left(x\right)={\left({\mathrm{sin}}^{2}\left(x\right)\right)}^{3}$
and ${\mathrm{sin}}^{2}\left(x\right)=1-{\mathrm{cos}}^{2}\left(x\right)$
$⇒{\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{6}\left(x\right)\mathrm{sin}\left(x\right)dx$
$={\int }_{0}^{\frac{\pi }{2}}{\left({\mathrm{sin}}^{2}\left(x\right)\right)}^{3}\mathrm{sin}xdx$
$={\int }_{0}^{\frac{\pi }{2}}{\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)}^{3}\mathrm{sin}xdx$
Apply u-substitution $u=\mathrm{cos}\left(x\right)⇒du=-\mathrm{sin}xdx$
$⇒-du=\mathrm{sin}xdx$
when $x=0,u=\mathrm{cos}\left(0\right)=1$
when $x=\frac{\pi }{2},u=\mathrm{cos}\left(\frac{\pi }{2}\right)=0$
Apply the formula ${\left(a-b\right)}^{3}={a}^{3}-{b}^{3}-3{a}^{2}b+3a{b}^{2}$
$={\int }_{0}^{1}\left[{\left({\left(1\right)}^{3}-\left({u}^{2}\right)\right)}^{3}-3{\left(1\right)}^{2}\left({u}^{2}\right)+3\left(1\right){\left({u}^{2}\right)}^{2}\right]du$
$={\int }_{0}^{1}\left(1-{u}^{6}-3{u}^{2}+3{u}^{4}\right)du$
$=\left(1-0\right)-\frac{1}{7}\left(1-0\right)+\frac{3}{5}\left(1-0\right)$
$=1-\frac{1}{7}-1+\frac{3}{5}=\frac{-1}{7}+\frac{3}{5}$