Integral exercise solved using Property of u-Substitution $\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left(a+b-x \right)dx}$

Exercise:

Compute, $\int_{0}^{2\pi }{\frac{x{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}$  where $n\in {{\mathbb{N}}^{*}}$

Solution: Let $I=\int_{0}^{2\pi }{\frac{x{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}$

Using the property $\int_{0}^{a}{f\left( x \right)dx}=\int_{0}^{a}{f\left( a-x \right)dx}$ to get:

$I=\int_{0}^{2\pi }{\frac{x{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}=\int_{0}^{2\pi }{\frac{\left( 2\pi -x \right){{\sin }^{2n}}\left( 2\pi -x \right)}{{{\sin }^{2n}}\left( 2\pi -x \right)+{{\cos }^{2n}}\left( 2\pi -x \right)}dx}$

   $=\int_{0}^{2\pi }{\frac{2\pi {{\sin }^{2n}}\left( 2\pi -x \right)}{{{\sin }^{2n}}\left( 2\pi -x \right)+{{\cos }^{2n}}\left( 2\pi -x \right)}}-\int_{0}^{2\pi }{\frac{x{{\sin }^{2n}}\left( 2\pi -x \right)}{{{\sin }^{2n}}\left( 2\pi -x \right)+{{\cos }^{2n}}\left( 2\pi -x \right)}dx}$

   $=2\pi \int_{0}^{2\pi }{\frac{{{\sin }^{2n}}\left( 2\pi -x \right)}{{{\sin }^{2n}}\left( 2\pi -x \right)+{{\cos }^{2n}}\left( 2\pi -x \right)}dx-I}$

$\Leftrightarrow 2I=2\pi \int_{0}^{2\pi }{\frac{{{\sin }^{2n}}\left( 2\pi -x \right)}{{{\sin }^{2n}}\left( 2\pi -x \right)+{{\cos }^{2n}}\left( 2\pi -x \right)}dx}$

$\Leftrightarrow I=\pi \int_{0}^{2\pi }{\frac{{{\sin }^{2n}}\left( 2\pi -x \right)}{{{\sin }^{2n}}\left( 2\pi -x \right)+{{\cos }^{2n}}\left( 2\pi -x \right)}dx}$

      $=\pi \int_{0}^{2\pi }{\frac{{{\sin }^{2n}}\left( 2\pi -x \right)}{{{\sin }^{2n}}\left( 2\pi -x \right)+{{\cos }^{2n}}\left( 2\pi -x \right)}dx}=\pi \int_{0}^{2\pi }{\frac{{{\left( -\sin x \right)}^{2n}}}{{{\left( -\sin x \right)}^{2n}}+{{\left( \cos x \right)}^{2n}}}dx}$

       $=\pi \int_{0}^{2\pi }{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}$

But $=\int_{0}^{2\pi }{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx=\int_{0}^{\frac{\pi }{2}}{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx+\int_{\frac{\pi }{2}}^{2\pi }{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}}}$

But ${{I}_{1}}=\int_{0}^{\frac{\pi }{2}}{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx=\int_{0}^{\frac{\pi }{2}}{\frac{{{\sin }^{2n}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{2n}}\left( \frac{\pi }{2}-x \right)+{{\cos }^{2n}}\left( \frac{\pi }{2}-x \right)}dx=\int_{0}^{\frac{\pi }{2}}{\frac{{{\cos }^{2n}}x}{{{\cos }^{2n}}x+{{\sin }^{2n}}x}dx}}}$

So $2{{I}_{1}}=\int_{0}^{\frac{\pi }{2}}{\frac{{{\sin }^{2n}}x+{{\cos }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx=\frac{\pi }{2}\Leftrightarrow {{I}_{1}}=\frac{\pi }{4}}$

Also ${{I}_{2}}=\int_{\frac{\pi }{2}}^{2\pi }{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx=}\int_{\frac{\pi }{2}}^{2\pi }{\frac{{{\sin }^{2n}}\left( \frac{5\pi }{2}-x \right)}{{{\sin }^{2n}}\left( \frac{5\pi }{2}-x \right)+{{\cos }^{2n}}\left( \frac{5\pi }{2}-x \right)}dx}$

$=\int_{\frac{\pi }{2}}^{2\pi }{\frac{{{\sin }^{2n}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{2n}}\left( \frac{\pi }{2}-x \right)+{{\cos }^{2n}}\left( \frac{\pi }{2}-x \right)}dx=\int_{\frac{\pi }{2}}^{2\pi }{\frac{{{\cos }^{2n}}x}{{{\cos }^{2n}}x+{{\sin }^{2n}}x}dx}}$

So $2{{I}_{2}}=\int_{\frac{\pi }{2}}^{2\pi }{\frac{{{\sin }^{2n}}x+{{\cos }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx=2\pi -\frac{\pi }{2}=\frac{3\pi }{2}}\Leftrightarrow {{I}_{2}}=\frac{3\pi }{4}$

Hence $\int_{0}^{2\pi }{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}={{I}_{1}}+{{I}_{2}}=\frac{\pi }{4}+\frac{3\pi }{4}=\pi$

Therefore $\int_{0}^{2\pi }{\frac{x{{\sin }^{2n}}x}{{{\cos }^{2n}}x+{{\sin }^{2n}}x}dx}={{\pi }^{2}}$