Mathematics

# Evaluate: $\displaystyle\int \dfrac { 1 } { 1 + \sin \theta } d \theta$

##### SOLUTION
$I=\displaystyle\int \dfrac{1}{1+\sin\theta}d\theta$

$I=\displaystyle\int \dfrac{1-\sin\theta}{(1+\sin\theta)(1-\sin\theta)}d\theta =\displaystyle\int\dfrac{1-\sin\theta}{1-\sin^2\theta}d\theta$

$I=\displaystyle\int \dfrac{1-\sin\theta}{\cos^2\theta}d\theta =\displaystyle\int (\sec^2\theta -\sec\theta \tan\theta)d\theta$

$I=\tan\theta -\sec \theta +c$.

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