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CGPET 2017 Solved Paper

Section: Mathematics

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Question : 101 of 150

Marks: +1, -0

Mathematics

\int\limits_{0}^{\pi/2} \sin 2x \tan^{-1}(\sin x) dx =

1
0
$\frac{\pi}{2}$
$\frac{\pi}{2} - 1$

Solution:

We have

I=\int\limits_{0}^{\pi/2} \sin 2x \tan^{-1}(\sin x) dx

I=\int\limits_{0}^{\pi/2} 2 \sin x \cos x \tan^{-1}(\sin x) dx

\text{Put} \sin x = t \Rightarrow \cos x \, dx = dt

Where,

x=0, t=0

x=\pi/2, t=1

\therefore I=\int\limits_{0}^{1} 2t \tan^{-1}(t) dt

\Rightarrow I=2\left[ \tan^{-1} t \cdot \frac{t^2}{2} \right]_{0}^{1}

-\int\limits_{0}^{1} \frac{1}{1+t^2} \cdot \frac{t^2}{2} \, dt ]

\Rightarrow I=2\left[ \left(\tan^{-1}1 \cdot \frac{1}{2}\right) - (0) \right.

-\frac{1}{2} \int\limits_{0}^{1} \left( \frac{t^2+1}{t^2+1} - \frac{1}{t^2+1} \right) dt

\Rightarrow I=2\left[ \frac{\pi}{8} - \frac{1}{2} \left[ t - \tan^{-1} t \right]_{0}^{1} \right]

\Rightarrow I=2\left[ \frac{\pi}{8} - \frac{1}{2} \left(1 - \frac{\pi}{4} - 0\right) \right]

\Rightarrow I=2\left[ \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} \right]

\Rightarrow I=2\left[ \frac{\pi}{4} - \frac{1}{2} \right]

\Rightarrow I = \frac{\pi}{2} - 1

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