∫ Sin X ( 1 + Cos X ) 2 D X - Mathematics

Sum
$\int\frac{\sin x}{\left( 1 + \cos x \right)^2} dx$

Solution

$\int\frac{\sin x}{\left( 1 + \cos x \right)^2}dx$
$\text{Let 1} + \cos x = t$
$\Rightarrow - \sin x = \frac{dt}{dx}$
$\Rightarrow \text{sin x dx} = - dt$
$Now, \int\frac{\sin x}{\left( 1 + \cos x \right)^2}dx$
$= \int - \frac{dt}{t^2}$
$= - \int t^{- 2} dt$
$= - \left[ \frac{t^{- 2 + 1}}{- 2 + 1} \right] + C$
$= \frac{1}{t} + C$
$= \frac{1}{1 + \cos x} + C$

Concept: Definite Integral as the Limit of a Sum
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.9 | Q 29 | Page 58