Mathematics

# Find :- $I = \displaystyle \int \dfrac{\sin \theta}{(4 \cos^2\theta)(2- \sin^2\theta)}d\theta$

##### SOLUTION
$\displaystyle I=\int \dfrac{\sin \theta d \theta }{(4\cos^{2} \theta )(2-\sin^{2} \theta )}$
Let $\cos \theta =t$
$-\sin \theta d \theta =dt$
$\sin \theta .d \theta =-dt$
$=\displaystyle -\int \dfrac{dt}{4t^{2}(2-1+t^{2})}$
$=\displaystyle -\int \dfrac{dt}{4t^{2}(1+t^{2})}$
$=\displaystyle -\dfrac{1}{4}\int \dfrac{1+t^{2}-t^{2}}{t^{2}(1+t^{2})}$
$=\displaystyle -\dfrac{1}{4}\int \left[\frac{1+t^{2}}{t^{2}(1+t^{2})}-\frac{t^{2}}{t^{2}(1+t^{2})}\right]dt$
$=\displaystyle -\dfrac{1}{4}\int \left[\frac{1}{t^{2}}-\frac{1}{1+t^{2}}\right]dt$
$=-\dfrac{1}{4}\left[-\dfrac{1}{t}-\tan^{-1}(t)\right]+C$
Put the value of $t$
$=\displaystyle -\dfrac{1}{4}\left[\dfrac{-1}{\cos\theta }-\tan^{-1}(\cos\theta )\right]+C$

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

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