Integral exercise asked in math Page Circulo Estudiantil Frederick Winslow Taylor

Exercise:

Evaluate, $\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx}\,\,\,,\,\,\,n\,\in \mathbb{N}\,$

Solution: we know that $\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx}=\int_{-\frac{\pi }{2}}^{0}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx+\int_{0}^{\frac{\pi }{2}}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx}}$

But $\int_{-\frac{\pi }{2}}^{0}{\frac{\cos x}{1+{{e}^{1-2n}}}dx}=-\int_{0}^{\frac{-\pi }{2}}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx}$

So $\int_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx}=-\int_{0}^{\frac{-\pi }{2}}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx+\int_{0}^{\frac{\pi }{2}}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx}}$

Let $u=-x\Leftrightarrow du=-dx$

So $-\int_{0}^{\frac{-\pi }{2}}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx}=\int_{0}^{\frac{\pi }{2}}{\frac{\cos \left( -u \right)}{1+{{e}^{-{{u}^{1-2n}}}}}du}$

Thus $\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx}=\int_{0}^{\frac{\pi }{2}}{\frac{\cos x}{1+{{e}^{{{x}^{1-2n}}}}}dx+\int_{0}^{\frac{\pi }{2}}{\frac{\cos x}{1+{{e}^{-{{x}^{1-2n}}}}}dx}}$

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

But $1+{{e}^{-{{x}^{1-2n}}}}=1+\frac{1}{{{e}^{{{x}^{1-2n}}}}}=\frac{{{e}^{{{x}^{1-2n}}}}+1}{{{e}^{{{x}^{1-2n}}}}}\Leftrightarrow \frac{1}{1+{{e}^{-{{x}^{1-2n}}}}}=\frac{{{e}^{{{x}^{1-2n}}}}}{1+{{e}^{{{x}^{1-2n}}}}}$

So $\frac{1}{1+{{e}^{{{x}^{1-2n}}}}}+\frac{1}{1+{{e}^{-{{x}^{1-2n}}}}}=\frac{1}{1+{{e}^{{{x}^{1-2n}}}}}+\frac{{{e}^{{{x}^{1-2n}}}}}{1+{{e}^{{{x}^{1-2n}}}}}=\frac{1+{{e}^{{{x}^{1-2n}}}}}{1+{{e}^{{{x}^{1-2n}}}}}=1$

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