Finding this integral using Half-Angle Identities

[Kimmy2]

I've been having trouble with this. Our teacher didn't explain this concept very well, and I don't have my textbook with me.

I started by changing the sin(96x) to (1/2)(1-cos(96*2x)).

For the cubed root of cosine, I tried (1/2)(1+cos(96*2)x)^(1/3)

From there, I got really confused on how to solve this integral. Help very much appreciated.

 

[soroban]

Hello, Kimmy2!

You're making the problem worse . . .


\(\displaystyle \text{We have: }\;\int \left[\cos(96)\right]^{\frac{1}{3}}\left[\sin(96x)\right]dx\)

\(\displaystyle \text{Let: }\:u \:=\:\cos(96x) \quad\Rightarrow\quad du \:=\:-96\sin(96x)\,dx \quad\Rightarrow\quad \sin(96x)\,dx \:=\:-\tfrac{1}{96}\,du\)

\(\displaystyle \text{Substitute: }\;\int u^{\frac{1}{3}}\left(-\tfrac{1}{96}\,du\right) \;=\;-\tfrac{1}{96}\int u^{\frac{1}{3}} du \;=\;-\tfrac{1}{96}\cdot\tfrac{3}{4}\,u^{\frac{4}{3}} + C \;=\;-\tfrac{1}{128}\,u^{\frac{4}{3}} + C\)

\(\displaystyle \text{Back-substitute: }\;-\tfrac{1}{128}\left[\cos(96x)\right]^{\frac{4}{3}} + C\)

 

[Kimmy2]

I see! That makes so much sense! Thank you so much for explaining this problem.

-Kimmy2