^{-1}≤ft[_{-π/2}^{π/2}{ x}{1+e^{x}}\,dx] =

Correct Answer is : π/4π/4π/4

Correct Answer is : π/4π/4π/4

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AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 77 of 160

Marks: +1, -0

\tan^{-1}\left[\int\limits_{-\pi/2}^{\pi/2}\frac{\cos x}{1+e^{x}}\,dx\right]

Solution:

Let us consider,

I = \int\limits_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{x}}\,dx

......(I)

Further simplify the above expression,

I = \int\limits_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{x}}\,dx

= \int\limits_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{x}}\,dx

= \int\limits_{-\pi/2}^{\pi/2} \frac{\cos\left(\frac{\pi}{2} - \frac{\pi}{2} - x\right)}{1 + e^{\frac{\pi}{2} - \frac{\pi}{2} - x}}\,dx

\left[\because \int\limits_{a}^{b} f(x)\,dx = \int\limits_{a}^{b} f(a+b-x)\,dx\right]

= \int\limits_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{-x}}\,dx

= \int\limits_{-\pi/2}^{\pi/2} \frac{e^{x}\cos x}{(1+e^{x})}\,dx

....(II)

Add equation (I) and (II).

2I = \int\limits_{-\pi/2}^{\pi/2} \frac{e^{x}\cos x}{(1+e^{x})}\,dx + \int\limits_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{x}}\,dx

\int\limits_{-\pi/2}^{\pi/2} \frac{(e^{x}+1)\cos x}{(1+e^{x})}\,dx

= \int\limits_{-\pi/2}^{\pi/2} \cos x\,dx

= \left[\sin x\right]_{-\pi/2}^{\pi/2}

Applying the limits, we get

2I = [1+1]

= 2

I = 1

Now the value of given function is,

\tan^{-1}\left[\int\limits_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{x}}\,dx\right] = \tan^{-1}(1)

= \tan^{-1}\left(\tan\frac{\pi}{4}\right)

= \pi/4

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