Thread: sub u w/ trig?

I have been getting better but I am having trouble with this evil integral.

Evalute the following integrel:

$\displaystyle \int_{}{} sin^2(3x+\frac{\pi}{8}) dx$

The answer is = $\displaystyle \frac{x}{2}- \frac{1}{12}sin(6x+\frac{\pi}{4})+c$


Here was my first path of deduction

$\displaystyle \int_{}{} sin^2(3x+\frac{\pi}{8}) dx$ ; $\displaystyle u=3x+\frac{\pi}{8}$ ,$\displaystyle u'=3$

dx in terms of u:$\displaystyle 1dx=3du \Rightarrow du=\frac{1}{3}$

$\displaystyle \int_{}{} sin^2(u) \frac{1}{3}du$ = $\displaystyle \frac{1}{3} \int_{}{} sin^2(u)du$ = $\displaystyle \frac{1}{3} \int_{}{} (sinu)^2du$

I am stuck here. I can't directly integrate anything.

Originally Posted by delgeezee

I have been getting better but I am having trouble with this evil integral.

Evalute the folling integrel:

$\displaystyle \int_{}{} sin^2(3x+\frac{\pi}{8}) dx$


Here was my first path of deduction

$\displaystyle \int_{}{} sin^2(3x+\frac{\pi}{8}) dx$ ; $\displaystyle u=3x+\frac{\pi}{8}$ ,$\displaystyle u'=3$

dx in terms of u:$\displaystyle 1dx=3du \Rightarrow du=\frac{1}{3}$

$\displaystyle \int_{}{} sin^2(u) \frac{1}{3}du$ = $\displaystyle \frac{1}{3} \int_{}{} sin^2(u)du$ = $\displaystyle \frac{1}{3} \int_{}{} (sinu)^2du$

I am stuck here. I can't directly integrate anything.

Hint: $\displaystyle \displaystyle \begin{align*} \sin^2{x} \equiv \frac{1}{2} - \frac{1}{2}\cos{2x} \end{align*} $