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TS EAMCET 18-July-2022 Shift 1 Solved Paper

Section: Mathematics

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Question : 77 of 160

Marks: +1, -0

Let

I=\int\limits_{-\pi/4}^{\pi/4} \frac{1}{2-\cos 2x} \left( \frac{3}{\pi} + \log \left( \frac{4+\sin x}{4-\sin x} \right) \right) dx

\int \frac{dx}{1+kx^2} = \frac{1}{\sqrt{k}} \tan^{-1}(\sqrt{k} x) + c, \tan^{-1}(0)=0

\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}

3 I^2 =

4
9
16
1

Solution: 👈: Video Solution

Let

I=\int\limits_{-\pi/4}^{\pi/4} \frac{1}{2-\cos 2x} \left( \frac{3}{\pi} + \log \left( \frac{4+\sin x}{4-\sin x} \right) \right) dx

Let

f(x) = \frac{1}{2-\cos 2x} \left( \frac{3}{\pi} + \log \left( \frac{4+\sin x}{4-\sin x} \right) \right)

f(-x) = \frac{1}{2-\cos 2x} \left( \frac{3}{\pi} + \log \left( \frac{4-\sin x}{4+\sin x} \right) \right)

= \frac{1}{2-\cos 2x} \left( \frac{3}{\pi} - \log \left( \frac{4+\sin x}{4-\sin x} \right) \right)

f(x)+f(-x) = \frac{6}{\pi(2-\cos 2x)}

I = \int\limits_{-\pi/4}^{\pi/4} f(x) dx = \int\limits_{0}^{\pi/4} f(x)+f(-x) dx

= \frac{6}{\pi} \int\limits_{0}^{\pi/4} \frac{1}{2-\cos 2x} dx

I = \frac{6}{\pi} \int\limits_{0}^{\pi/4} \frac{1}{2 - \left( \frac{1-\tan^2 x}{1+\tan^2 x} \right)} dx

= \frac{6}{\pi} \int\limits_{0}^{\pi/4} \frac{1+\tan^2 x}{1+3\tan^2 x} dx

= \frac{6}{\pi} \int\limits_{0}^{\pi/4} \frac{\sec^2 x}{1+3\tan^2 x} dx

On putting

\tan x = t

I = \frac{6}{\pi} \int\limits_{0}^{1} \frac{\sec^2 x \, dx = dt}{1+(\sqrt{3} t)^2}

= \frac{6}{\pi} \cdot \frac{1}{\sqrt{3}} \left[ \tan^{-1}(\sqrt{3} t) \right]_{0}^{1}

= \frac{2\sqrt{3}}{\pi} \left[ \tan^{-1}(\sqrt{3}) - 0 \right]

I = \frac{2\sqrt{3}}{\pi} \times \frac{\pi}{3} = \frac{2}{\sqrt{3}}

\Rightarrow I^{2} = \frac{4}{3}

3 I^{2} = 4

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