Aine Sellers

2022-02-25

I am trying to find this by using integration by parts but I am not sure how to do it.
${\int }_{0}^{\frac{\pi }{2}}{\left(\mathrm{sin}x\right)}^{7}{\left(\mathrm{cos}x\right)}^{5}dx$

nastaja1en

To integrate the function
$f\left(x\right)={\mathrm{sin}}^{n}x\cdot {\mathrm{cos}}^{m}x$,
when n or m are positive odd numbers, we can apply a general technique which consists of expanding f(x) into a sum of terms of the form

or

Using the identity
${\mathrm{cos}}^{2}x=1-{\mathrm{sin}}^{2}x$
in the form
${\mathrm{cos}}^{4}x={\left(1-{\mathrm{sin}}^{2}x\right)}^{2}=1-2{\mathrm{sin}}^{2}x+{\mathrm{sin}}^{4}x$,
we rewrite our
$f\left(x\right)={\mathrm{sin}}^{7}x\cdot {\mathrm{cos}}^{5}x={\mathrm{sin}}^{7}x\cdot {\mathrm{cos}}^{4}x\cdot \mathrm{cos}x$ as
as
$f\left(x\right)={\mathrm{sin}}^{7}x\cdot \left(1-2{\mathrm{sin}}^{2}x+{\mathrm{sin}}^{4}x\right)\cdot \mathrm{cos}x$
$={\mathrm{sin}}^{7}x\cdot \mathrm{cos}x-2{\mathrm{sin}}^{9}x\cdot \mathrm{cos}x+{\mathrm{sin}}^{11}x\cdot \mathrm{cos}x$
Each term is of the form ${\mathrm{sin}}^{p}x\cdot \mathrm{cos}x$ and can easily be integrated by the substitution :
$\int {\mathrm{sin}}^{p}x\cdot \mathrm{cos}xdx=\int {u}^{p}du=\frac{{u}^{p+1}}{p+1}=\frac{{\mathrm{sin}}^{p+1}x}{p+1}+C$
${\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{p}x\cdot \mathrm{cos}xdx=\frac{1}{p+1}$
${\int }_{0}^{\frac{\pi }{2}}f\left(x\right)dx={\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{7}x{\mathrm{cos}}^{5}xdx$

shotokan0758s

Let $I={\int }_{0}^{\frac{\pi }{2}}{\left(\mathrm{sin}x\right)}^{7}{\left(\mathrm{cos}x\right)}^{5}dx$
Then using the substitution $u=\frac{\pi }{2}-x$ we have $I={\int }_{0}^{\frac{\pi }{2}}{\left(\mathrm{sin}x\right)}^{5}{\left(\mathrm{cos}x\right)}^{7}dx$
$2I={\int }_{0}^{\frac{\pi }{2}}{\left(\mathrm{sin}x\right)}^{5}{\left(\mathrm{cos}x\right)}^{5}dx$
Since ${\mathrm{cos}}^{2}\left(x\right)+{\mathrm{sin}}^{2}\left(x\right)=1$
And that
$2I=\frac{1}{{2}^{5}}{\int }_{0}^{\frac{\pi }{2}}{\left(\mathrm{sin}2x\right)}^{5}dx$
since $\mathrm{sin}\left(2x\right)=2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)$. Then substituting $u=2x$ gives
$2I=\frac{1}{{2}^{6}}{\int }_{0}^{\pi }{\left(\mathrm{sin}u\right)}^{5}du=\frac{1}{{2}^{6}}{\int }_{0}^{\pi }\left(\mathrm{sin}u\right){\left(1-{\mathrm{cos}}^{2}u\right)}^{2}du$
$=\frac{1}{{2}^{6}}{\int }_{0}^{\pi }\left(\mathrm{sin}u\right)\left(1-2{\mathrm{cos}}^{2}u+{\mathrm{cos}}^{4}\left(u\right)\right)du$
Hence
$I=\frac{1}{{2}^{7}}{\left[-\mathrm{cos}u+\frac{2{\mathrm{cos}}^{3}u}{3}-\frac{{\mathrm{cos}}^{5}u}{5}\right]}_{0}^{\pi }$
$I=\frac{1}{{2}^{7}}{\left[2+\frac{-2\cdot 2}{3}+\frac{2}{5}\right]}_{0}^{\pi }$
Hence
$I=\frac{1}{{2}^{3}\cdot 15}=\frac{1}{120}$

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