\(\tan ^{-1}\left[\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x\right]=\)
\(\tan ^{-1}\left[\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x\right]=\)
\(\frac{\pi}{4}\)
\(\frac{\pi}{3}\)
\(\frac{\pi}{6}\)
\(\frac{\pi}{2}\)
Solution
Given, \(\tan ^{-1}\left[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x\right]\)
Let \(I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x\)...(i)
\(\begin{aligned}
& \Rightarrow \quad I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-\frac{\pi}{2}-x\right)}{\left.1+e^{\left(\frac{\pi}{2}-\frac{\pi}{2}-x\right.}\right)} d x \\
& {\left[\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]} \\
& =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^{-x}} d x=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^x \cos x}{1+e^x} d x \quad \ldots (ii)
\end{aligned}\)
Adding Eqs. (i) and (ii), we get
\(\begin{gathered}
2 I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(e^x+1\right) \cos x}{\left(1+e^x\right)} d x \quad \ldots (iii) \\
=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x d x=[\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\
\Rightarrow 2 I=[1+1]=2 \Rightarrow I=1 \\
\therefore \tan ^{-1}\left[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x\right]=\tan ^{-1}(1)=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4}
\end{gathered}\)