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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 76 of 160

Marks: +1, -0

\int\limits_{0}^{\pi/2} \frac{dx}{4+5 \sin x} =

$\frac{1}{2} \log 3$
$\frac{1}{3} \log 2$
$2 \log 3$
$\frac{1}{2} \log \frac{3}{2}$

Solution:

We have,

\int\limits_{0}^{\pi/2} \; \frac{dx}{4+5 \sin x}

= \int\limits_{0}^{\pi/2} \; \frac{dx}{4+5\left(\frac{2 \tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}\right)}

= \int\limits_{0}^{\pi/2} \; \frac{1+\tan^2 \frac{x}{2}}{4+4 \tan^2 \frac{x}{2}+10 \tan \frac{x}{2}} dx

= \; \frac{1}{4} \int\limits_{0}^{\pi/2} \; \frac{\sec^2 \frac{x}{2}}{\tan^2 \frac{x}{2}+\frac{5}{2} \tan \frac{x}{2}+1} dx

let

\tan \frac{x}{2} = t

\Rightarrow \; \sec^2 \frac{x}{2} \left(\frac{1}{2}\right) dx = dt

\Rightarrow \sec^2 \frac{x}{2} dx = 2 dt

at

x=0, t=0

and

\; \; x = \frac{\pi}{2}, t=1 = \frac{2}{4} \int\limits_{0}^{1} -\frac{1}{t^2+\frac{5}{2}t+1} dt

= \; \frac{1}{2} \int\limits_{0}^{1} \; \frac{dt}{t^2+\frac{5}{2}t+1+\frac{25}{16}-\frac{25}{16}}

= \; \frac{1}{2} \int\limits_{0}^{1} \; \frac{dt}{\left(t+\frac{5}{4}\right)^2 - \frac{9}{16}} = \; \frac{1}{2} \int\limits_{0}^{1} \; \frac{dt}{\left(t+\frac{5}{4}\right)^2 - \left(\frac{3}{4}\right)^2}

= \; \frac{1}{3} \left[ \log \left| \frac{t+\frac{1}{2}}{t+2} \right| \right]_{0}^{1}

= \; \frac{1}{3} \left[ \log \left( \frac{\frac{3}{2}}{3} \right) - \log \left( \frac{\frac{1}{2}}{2} \right) \right]

= \; \frac{1}{3} \left[ \log \left( \frac{1}{2} \right) - \log \left( \frac{1}{4} \right) \right] = \; \frac{1}{3} \left[ \log \left( \frac{1}{\frac{2}{1}4} \right) \right]

= \; \frac{1}{3} \log 2

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