MHB Definite integral challenge ∫cos2017xsin2017xdx

[lfdahl]

Calculate the following definite trigonometric integral:

\[\int_{0}^{\frac{\pi}{2}} \cos^{2017}x \sin^{2017}x dx\].

 

[Albert1]

Calculate the following definite trigonometric integral:

\[\int_{0}^{\frac{\pi}{2}} \cos^{2017}x \sin^{2017}x dx\]=A.

my solution :

Spoiler

$Using \,Beta \,Function :$
$$\beta(m,n)=2\int_{0}^{\frac{\pi}{2}}sin^{2m-1}x\,\,cos^{2n-1}x\, dx
=\dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=\dfrac{(m-1)!(n-1)!}{(m+n-1)!}$$
here $m=n=1009$
so $A=\dfrac{\Gamma(1009)\Gamma(1009)}{2\Gamma(2018)}=\dfrac{(1008)!\times(1008)!}{2\times(2017)!}$

 

[lfdahl]

my solution :

Spoiler

$Using \,Beta \,Function :$
$$\beta(m,n)=2\int_{0}^{\frac{\pi}{2}}sin^{2m-1}x\,\,cos^{2n-1}x\, dx
=\dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=\dfrac{(m-1)!(n-1)!}{(m+n-1)!}$$
here $m=n=1009$
so $A=\dfrac{\Gamma(1009)\Gamma(1009)}{2\Gamma(2018)}=\dfrac{(1008)!\times(1008)!}{2\times(2017)!}$

Thankyou, Albert!, for your fine solution! Well done.

 

[lfdahl]

An alternative solution:

Spoiler

\[\int_{0}^{\frac{\pi}{2}}\cos^{2017}x \sin^{2017}x dx =2^{-2017}\int_{0}^{\frac{\pi}{2}} \sin^{2017}2x dx\]Consider the general case: $\int_{0}^{\frac{\pi}{2}} \sin^{n}2x dx$.

Integration by parts:
\[\int_{0}^{\frac{\pi}{2}} \sin^{n}2x dx = \int_{0}^{\frac{\pi}{2}} \sin 2x\sin^{n-1}2x dx \\\\ =\left [ -\frac{\cos 2x}{2} \sin^{n-1}2x \right ]_0^{\frac{\pi}{2}}+(n-1)\int_{0}^{\frac{\pi}{2}} \cos^2 2x\sin^{n-2}2xdx \\\\ = (n-1)\int_{0}^{\frac{\pi}{2}} \left ( 1-\sin^2 2x\right ) \sin^{n-2}2xdx \\\\ = (n-1)\int_{0}^{\frac{\pi}{2}} \sin^{n-2}2xdx-(n-1)\int_{0}^{\frac{\pi}{2}} \sin^{n}2xdx\]
- Thus we arrive at the reduction form:

\[\Rightarrow \int_{0}^{\frac{\pi}{2}} \sin^{n}2xdx = \frac{n-1}{n}\int_{0}^{\frac{\pi}{2}} \sin^{n-2}2xdx\]

Now, we use this equation repeatedly:

\[\Rightarrow \int_{0}^{\frac{\pi}{2}} \sin^{n}2xdx = \frac{n-1}{n}\cdot \frac{n-3}{n-2}\int_{0}^{\frac{\pi}{2}} sin^{n-4 }2xdx \\\\ \\\\= \left\{\begin{matrix} \frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot ...\cdot \frac{2}{3}\int_{0}^{\frac{\pi}{2}} \sin 2xdx, \: \: \: n\: \: odd\\ \\ \frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot ...\cdot \frac{3}{4}\cdot \frac{1}{2}\int_{0}^{\frac{\pi}{2}} dx, \: \: \: n\: \: even \end{matrix}\right. \\\\ \\\\ =\left\{\begin{matrix} \frac{(n-1)!}{n!}, \: \: \: n\: \: odd\\ \\ \frac{(n-1)!}{n!}\frac{\pi}{2} , \: \: \: n\: \: even\end{matrix}\right.\]Returning to our starting problem, we get:

\[\int_{0}^{\frac{\pi}{2}}\cos^{2017}x \sin^{2017}x dx =2^{-2017}\int_{0}^{\frac{\pi}{2}} \sin^{2017}2x dx =2^{-2017}\frac{2016!}{2017!}
\\\\= \frac{1}{2}\cdot 2^{-2016}\frac{(2016!)^2}{2017!}=\frac{(1008!)^2}{2\cdot 2017!}\]