Mathematics - Definite Integration Question with Solution | TestHub

Let $I={\int }_{-2\pi }^{2\pi }\frac{{\sin }^{6}x}{\left({\sin }^{6}x+{\cos }^{6}x\right)\left(1+e^{-x}\right)}dx$.

Using the property ${\int }_{-a}^{a}f(x)dx={\int }_0^a(f(x)+f(-x))dx$.

Let $f(x)=\frac{{\sin }^{6}x}{\left({\sin }^{6}x+{\cos }^{6}x\right)\left(1+e^{-x}\right)}$.

Then $f(-x)=\frac{{\sin }^6(-x)}{\left({\sin }^6(-x)+{\cos }^6(-x)\right)\left(1+e^{-(-x)}\right)}=\frac{{\sin }^{6}x}{\left({\sin }^{6}x+{\cos }^{6}x\right)\left(1+e^x\right)}$.

$f(x)+f(-x)=\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}\left(\frac{1}{1+e^{-x}}+\frac{1}{1+e^x}\right)$

$=\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}\left(\frac{e^x}{e^x+1}+\frac{1}{1+e^x}\right)$

$=\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}\left(\frac{e^x+1}{e^x+1}\right)=\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}$.

So, $I={\int }_0^{2\pi }\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}dx$.

Using the property ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$.

Let $J={\int }_0^{2\pi }\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}dx$.

Then $J={\int }_0^{2\pi }\frac{{\sin }^6(2\pi -x)}{{\sin }^6(2\pi -x)+{\cos }^6(2\pi -x)}dx={\int }_0^{2\pi }\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}dx$.

This property doesn't simplify it further.

However, we can use the property ${\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx$ if $f(2a-x)=f(x)$.

Here, $f(x)=\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}$.

$f(2\pi -x)=\frac{{\sin }^6(2\pi -x)}{{\sin }^6(2\pi -x)+{\cos }^6(2\pi -x)}=\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}=f(x)$.

So $I=2{\int }_0^{\pi }\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}dx$.

Now use ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$.

Let $K={\int }_0^{\pi }\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}dx$.

$K={\int }_0^{\pi }\frac{{\sin }^6(\pi -x)}{{\sin }^6(\pi -x)+{\cos }^6(\pi -x)}dx={\int }_0^{\pi }\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}dx$.

This property doesn't simplify it further.

Let's use a different approach for $J={\int }_0^{2\pi }\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}dx$.

Let $J={\int }_0^{2\pi }\frac{{\sin }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}dx$.

Also, $J={\int }_0^{2\pi }\frac{{\cos }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}dx$ (by replacing $x$ with $\frac{\pi }{2}-x$ and considering the periodicity, or by symmetry).

Adding these two forms:

$2J={\int }_0^{2\pi }\frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{6}x+{\cos }^{6}x}dx={\int }_0^{2\pi }1dx=2\pi$.

So $2J=2\pi$, which means $J=\pi$.

Therefore, $I=\pi$.

The final answer is $\boxed{\pi }$.