Q117 / 160 · AP EAMCET 2019 (22 Apr Shift 1) · Mathematics

\(\tan ^{-1}\left[\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x\right]=\)

  • Correct answer
    A
    \(\frac{\pi}{4}\)
  • B
    \(\frac{\pi}{3}\)
  • C
    \(\frac{\pi}{6}\)
  • D
    \(\frac{\pi}{2}\)

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 \(\begin{gathered} 2 I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(e^x+1\right) \cos x}{\left(1+e^x\right)} d x \quad \ldots (iii) \\ =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x d x=[\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\ \Rightarrow 2 I=[1+1]=2 \Rightarrow I=1 \\ \therefore \tan ^{-1}\left[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x\right]=\tan ^{-1}(1)=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4} \end{gathered}\)

Hence, the correct answer is option: 'A'.

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