Mathematics - Definite Integration Question with Solution | TestHub

Using the property ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$ with $a=\pi /2$, we have:$$I={\int }_0^{\pi /2}\frac{{\sin }^2(\pi /2-x)}{e^{\pi /2-x}+e^{\pi /2-(\pi /2-x)}}dx={\int }_0^{\pi /2}\frac{{\cos }^{2}x}{e^{\pi /2-x}+e^x}dx$$ Adding the original integral and this new form:$$2I={\int }_0^{\pi /2}\frac{{\sin }^{2}x+{\cos }^{2}x}{e^x+e^{\pi /2-x}}dx={\int }_0^{\pi /2}\frac{1}{e^x+e^{\pi /2-x}}dx$$ Multiply numerator and denominator by $e^x$:$$2I={\int }_0^{\pi /2}\frac{e^x}{e^{2x}+e^{\pi /2}}dx$$ Let $u=e^x$, so $du=e^{x}dx$. When $x=0,u=1$. When $x=\pi /2,u=e^{\pi /2}$.$$2I={\int }_1^{e^{\pi /2}}\frac{du}{u^2+(e^{\pi /4}{)}^2}=\frac{1}{e^{\pi /4}}{\left[{\tan }^{-1}\left(\frac{u}{e^{\pi /4}}\right)\right]}_1^{e^{\pi /2}}$$ $$2I=e^{-\pi /4}\left[{\tan }^{-1}(e^{\pi /4})-{\tan }^{-1}(e^{-\pi /4})\right]$$ Using the identity ${\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)$ or ${\tan }^{-1}x+{\tan }^{-1}(1/x)=\pi /2$ for $x>0$:$$2I=e^{-\pi /4}\left(2{\tan }^{-1}(e^{\pi /4})-\frac{\pi }{2}\right)$$ $$I=\frac{1}{2}{e}^{-\pi /4}\left(2{\tan }^{-1}(e^{\pi /4})-\frac{\pi }{2}\right)\approx 0.1625$$ Rounding to two decimal places, $I\approx 0.16$.