Mathematics

# Solve: $\displaystyle \int \dfrac{\cos x}{(a+b\sin x)^2}dx$

##### SOLUTION
Now,
$\displaystyle \int \dfrac{\cos x}{(a+b\sin x)^2}dx$
$=\dfrac{1}{b}\displaystyle \int \dfrac{b\cos x}{(a+b\sin x)^2}dx$
$=\dfrac{1}{b}\displaystyle \int \dfrac{d(a+b\sin x)}{(a+b\sin x)^2}$
$=-\dfrac{1}{b}.\dfrac{1}{a+b\sin x }+c$ [ Where $c$ is integrating constant]

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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