How do you integrate $\int \frac{sin x}{{(2 + 3 cos x)}^{2}}$ $int sinx/(2+3cosx)^2$ using substitution?

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Answer 1 · 2018-03-14T11:58:33

The answer is $= \frac{1}{3} \cdot (\frac{1}{2 + 3 cos x}) + C$ $=1/3*(1/(2+3cosx))+C$

Let $u = 2 + 3 cos x$ $u=2+3cosx$

$d u = - 3 sin x d x$ $du=-3sinxdx$

$sin x d x = - \frac{1}{3} d u$ $sinxdx=-1/3du$

Therefore,

$\int \frac{sin x d x}{{(2 + 3 cos x)}^{2}} = - \frac{1}{3} \int \frac{d u}{u^{2}}$ $int(sinxdx)/(2+3cosx)^2=-1/3int(du)/(u^2)$

$= - \frac{1}{3} \int u^{- 2} d u = \frac{1}{3} \cdot \frac{1}{u}$ $=-1/3intu^-2du=1/3*1/u$

$= \frac{1}{3} \cdot (\frac{1}{2 + 3 cos x}) + C$ $=1/3*(1/(2+3cosx))+C$

How do you integrate ∫sinx(2+3cosx)2int sinx/(2+3cosx)^2 using substitution?