sub u w/ trig?

**delgeezee** · Dec 4th 2011, 07:50 PM

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

Evalute the following integrel:

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

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

Here was my first path of deduction

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

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

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

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

**Prove It** · Dec 4th 2011, 08:08 PM

Originally Posted by delgeezee

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

Evalute the folling integrel:

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

Here was my first path of deduction

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

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

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

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

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

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