Integration with trig.

[roxstar1]

could someone please point me in the right direction here?


integral (1/(2-3sinx)) dx


I think Im supposed to use a triangle substitution here. I have found sinx = 2u/(1+u^2)...I am having problems finding dx in terms of u however, which is making the integral difficult to solve. Any suggestions would be much appreciated.

 

[soroban]

Hello, roxstar1!

\(\displaystyle \L\int \frac{dx}{2\,-\,3\sin x}\)

I think Im supposed to use a triangle substitution here.
I have found \(\displaystyle \sin x\:=\:\frac{2u}{1\,+\,u^2}\;\;\) . . . Yes!
I am having problems finding \(\displaystyle dx\) in terms of \(\displaystyle u\).

This substitution is based on: \(\displaystyle \,u\:=\:\tan\left(\frac{x}{2}\right)\)

After a lot of calculus, trig, and algebra, we get these three formula:

\(\displaystyle \L\;\;\sin x\:=\:\frac{2u}{1\,+\,u^2}\;\;\;\;\cos x\:=\:\frac{1\,-\,u^2}{1\,+\,u^2}\;\;\;\;dx\:=\:\frac{2\,du}{1\,+\,u^2}\)


I recommend memorizing them, rather than deriving them each time.
If you'd like to see the derivations, please let me know.

 

[roxstar1]

thanks! that makes the problem a bit easier