MHB Evaluating Integrals Involving Trig Functions

[sbhatnagar]

Evaluate:

1. $\displaystyle \int_0^{\displaystyle 2\pi} \frac{x \sin^{2n}(x)}{\sin^{2n}(x)+\cos^{2n}(x)}dx$, $n>0$

2. $\displaystyle \int_0^{ \displaystyle \pi \over \displaystyle 2} \frac{x \sin x \cos x}{\sin^{4}(x)+\cos^{4}(x)}dx$

 

[Mathwebster]

In 1) - is $n$ a positive integer?

 

[sbhatnagar]

In 1) - is $n$ a positive integer?

Yes! Sorry, I should I have mentioned at the beginning.

 

[sbhatnagar]

Here is a hint: Try using properties of definite integrals.

 

[Sherlock1]

Putting $x \mapsto 2\pi-x$ we have:$\displaystyle I = \int_{0}^{2\pi}\frac{x\sin^{2n}{x}}{\sin^{2n}{x}+ \cos^{2n}{x}}\;{dx} = \int_{0}^{2\pi}\frac{(2\pi-x)\sin^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx} = 2\pi\int_{0}^{2\pi}\frac{\sin^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx}-I$.Thus $ \displaystyle I = \pi\int_{0}^{2\pi}\frac{\sin^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx} $. Now, write this from $[0, ~ \pi/2]$ and $[\pi/2, ~ 2\pi]$, then:$\displaystyle I = \pi\int_{0}^{\pi/2}\frac{\sin^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx} +\pi\int_{\pi/2}^{2\pi}\frac{\sin^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx}$. Put $x\mapsto \frac{\pi}{2}-x$ and $x\mapsto \frac{5\pi}{2}-x$, then:$\displaystyle I = \pi\int_{0}^{\pi/2}\frac{\cos^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx} +\pi\int_{\pi/2}^{2\pi}\frac{\cos^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx}$. Add this and the above, then we've:

$\displaystyle 2I = 2\pi^2 \Rightarrow \boxed{I = \pi^2}$. I'll leave the second one for others to try.

 

[sbhatnagar]

Putting $x \mapsto 2\pi-x$ we have:$\displaystyle I = \int_{0}^{2\pi}\frac{x\sin^{2n}{x}}{\sin^{2n}{x}+ \cos^{2n}{x}}\;{dx} = \int_{0}^{2\pi}\frac{(2\pi-x)\sin^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx} = 2\pi\int_{0}^{2\pi}\frac{\sin^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx}-I$.Thus $ \displaystyle I = \pi\int_{0}^{2\pi}\frac{\sin^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx} $. Now, write this from $[0, ~ \pi/2]$ and $[\pi/2, ~ 2\pi]$, then:$\displaystyle I = \pi\int_{0}^{\pi/2}\frac{\sin^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx} +\pi\int_{\pi/2}^{2\pi}\frac{\sin^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx}$. Put $x\mapsto \frac{\pi}{2}-x$ and $x\mapsto \frac{5\pi}{2}-x$, then:$\displaystyle I = \pi\int_{0}^{\pi/2}\frac{\cos^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx} +\pi\int_{\pi/2}^{2\pi}\frac{\cos^{2n}{x}}{\sin^{2n}{x}+\cos^{2n}{x}}\;{dx}$. Add this and the above, then we've:

$\displaystyle 2I = 2\pi^2 \Rightarrow \boxed{I = \pi^2}$. I'll leave the second one for others to try.

Yeah, that's right.

 

[sbhatnagar]

Solution to 2.

Let $\displaystyle I= \int_0^{\pi / 2}\frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx$

By a property of definite integrals, we get

$\displaystyle I= \int_0^{\pi /2}\dfrac{\displaystyle \left( \dfrac{\pi}{2} -x\right) \sin \left( \dfrac{\pi}{2} -x\right) \cos \left( \dfrac{\pi}{2} -x\right)}{ \sin^{4}\left( \dfrac{\pi}{2} -x\right) + \cos^4 \left( \dfrac{\pi}{2} -x\right)}dx = \int_0^{\pi /2}\dfrac{\displaystyle \left( \dfrac{\pi}{2} -x\right) \sin x \cos x}{ \sin^{4}x+ \cos^4 x }dx ={\pi \over 2}\int_0^{\pi / 2}\frac{ \sin x \cos x}{\sin^4 x + \cos^4 x}dx -I$

$\displaystyle \therefore \ \ 2I= \frac{\pi}{2} \int_0^{\pi/2}\frac{\sin x \cos x}{\sin^4 x + \cos ^4 x}dx =\frac{\pi}{2} \int_0^{\pi /2}\frac{\tan x \sec^2 x}{1+\tan^4 x }dx$

Substitute $t= \tan^2 x$ and $dt = 2 \tan x \sec^2 x \ dx$.

$\displaystyle 2I= \frac{\pi}{4}\int_{0}^{\infty}\frac{1}{1+t^2}dt = \frac{\pi^2}{8}$

$\displaystyle \therefore \ \ I= \frac{\pi^2}{16}$