int frac{sin ^{2} x}{1+cos x} d x | vedclass

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Question

(B) Given the integral $ I = \int \frac{\sin ^{2} x}{1+\cos x} d x $. Using the trigonometric identity $ \sin ^{2} x = 1 - \cos ^{2} x $,we get: $ I =

Vedclass · Accepted Answer

$ x-\sin x+C $

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(B) Given the integral $ I = \int \frac{\sin ^{2} x}{1+\cos x} d x $. Using the trigonometric identity $ \sin ^{2} x = 1 - \cos ^{2} x $,we get: $ I = \int \frac{1 - \cos ^{2} x}{1 + \cos x} d x $. Since $ 1 - \cos ^{2} x = (1 - \cos x)(1 + \cos x) $,the expression becomes: $ I = \int \frac{(1 - \cos x)(1 + \cos x)}{1 + \cos x} d x $. Canceling the common term $ (1 + \cos x) $,we have: $ I = \int (1 - \cos x) d x $. Integrating term by term,we get: $ I = x - \sin x + C $.

$ \int \frac{\sin ^{2} x}{1+\cos x} d x $

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