Mathematics

# Evaluate : $\displaystyle \int\frac{\cos x}{\sin(x-\dfrac{\pi}{6})\sin(x+\dfrac{\pi}{6})}dx$

$\log \left | \displaystyle \frac {2\sin x-1}{2\sin x+1} \right |+C$

##### SOLUTION
$\int\dfrac{\cos x}{\sin(x-\dfrac{\pi}{6})\sin(x+\dfrac{\pi}{6})}dx$

$\int\dfrac{\cos x}{\sin^2x-\sin^2\dfrac {\pi}{6}}dx$

Let $\sin x=t$

or $dt= \cos x dx$

$\therefore I=\int\dfrac {dt}{t^2-\dfrac {1}{4}}$

$=\dfrac {1}{2 \dfrac{1}{2}}\log \left | \dfrac {t-\dfrac{1}{2}}{t+\dfrac{1}{2}}\right |+C$

$=\log \left | \dfrac {2t-1}{2t+1} \right |+C$

$=\log \left | \dfrac {2\sin x-1}{2\sin x+1} \right |+C$  where $t=\sin x$

Hence, option 'A' is correct.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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