sin-2-x-dx-1-cos-x-2-

Question Number 92960 by i jagooll last updated on 10/May/20

\int \frac{\sin^{2} x dx}{{(1 + \cos x)}^{2}}

Answered by i jagooll last updated on 10/May/20

\frac{1 - \cos^{2} x}{{(1 + \cos x)}^{2}} = \frac{1 - \cos x}{1 + \cos x} = \tan^{2} (\frac{x}{2})

= \sec^{2} (\frac{x}{2}) - 1

\int \frac{\sin^{2} x dx}{{(1 + \cos x)}^{2}} = \int [\sec^{2} (\frac{x}{2}) - 1] dx

= 2 \tan (\frac{x}{2}) - x + c

Commented by john santu last updated on 10/May/20

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Answered by $@ty@m123 last updated on 10/May/20

A l t e r n a t i v e w a y :

\frac{\sin^{2} x}{{(1 + \cos x)}^{2}}

= \frac{1 - \cos x}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x}

= {(\frac{1 - \cos x}{\sin x})}^{2}

= {(cosec x - \cot x)}^{2}

= cosec^{2} x + \cot^{2} x - 2 cosec x \cot x

= 2 cosec^{2} x - 1 - 2 cosec x \cot x

∴ \int (2 cosec^{2} x - 1 - 2 cosec x \cot x) d x

= - 2 \cot x - x + cosec x + C