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

Question

4272 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-14T11:58:33

The answer is $=1/3*(1/(2+3cosx))+C$

Let $u=2+3cosx$

$du=-3sinxdx$

$sinxdx=-1/3du$

Therefore,

$int(sinxdx)/(2+3cosx)^2=-1/3int(du)/(u^2)$

$=-1/3intu^-2du=1/3*1/u$

$=1/3*(1/(2+3cosx))+C$

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