# Trig integrals

• May 8th 2013, 11:56 AM
Lozeee
Trig integrals
h(theta)=2sin(3theta^2)/Cos^4(2theta)

i use substitution
letting u=3theta^2

and i keep getting 1/6 - 1/6Cos6theta

• May 8th 2013, 01:02 PM
HallsofIvy
Re: Trig integrals
"u= 3theta" isn't going to help you with "2theta". Instead use trig identities to reduce sin(3theta) and cos(2theta) to sin(theta) and cos(\theta)
cos(2theta)= cos^2(theta)- sin^2(theta) you should know "by heart".
sin(3theta) is more complicated. sin(3theta)= sin(2theta+ theta)= sin(2theta)cos(theta)+ cos(2theta)sin(theta)= (2cos(theta)sin(theta))cos(theta)+ (cos^2(theta)- sin^2(theta))sin(theta)= 3cos^2(theta)sin(theta)- sin^2(theta)

(If that is really "sin(3theta^2)" rather than "sin^2(3theta)", that integral cannot be done in terms of elementary functions.)
• May 8th 2013, 05:02 PM
Prove It
Re: Trig integrals
Quote:

Originally Posted by Lozeee
h(theta)=2sin(3theta^2)/Cos^4(2theta)

i use substitution
letting u=3theta^2

and i keep getting 1/6 - 1/6Cos6theta

Am I correct in assuming that your function is actually \displaystyle \displaystyle \begin{align*} \int{\frac{2\sin^2{(3\theta)}}{\cos^4{(2\theta)}}\ ,d\theta} \end{align*}?