Mathematics

# Solve $\displaystyle\int { \dfrac { \sin { x } }{ 1-4\cos ^{ 2 }{ x } } } .dx.$

##### SOLUTION
$\displaystyle\int { \dfrac { \sin { x } }{ 1-4\cos ^{ 2 }{ x } } } .dx.$

Put, $2\cos x = t$

$\implies - 2 \sin x dx = dt$

Now, $\displaystyle\int \dfrac{\sin x}{1 - 4\cos^2 x} dx = \dfrac{1}{2} \int \dfrac{-1}{1 - t^2} dt$

$= \dfrac{1}{2} \displaystyle\int \dfrac{1}{t^2 - 1} dt$

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

$= \dfrac{1}{4} \log \left|{\dfrac{2\cos x - 1}{2\cos x + 1}} \right|+ c$ (Ans)

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

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