Trigonometric Integrals

• Feb 5th 2009, 05:33 PM
Craig22
Trigonometric Integrals
Okay, I am having major problems using the reduction formulas for sine and cosine. I look at the examples in the book, and I can't figure out where some of the numbers come from. I have a problem if anyone can show me how to do it...

$
fsin^4xcos^2x dx
$

I've gotten this far:

$fsin^4x(1-sin^2x)dx$

$fsin^4x dx - fsin^6x dx$

I am not sure how to proceed from this point. Any help is appreciated!
• Feb 5th 2009, 09:03 PM
Abu-Khalil
Let $I_n=\int\sin^n xdx$, then using partial integration $I_6=\int(-\cos x)'\sin^5xdx=-\cos x\sin ^5 x + 5\int \sin^4 x\cos ^2 x dx$ but $\int \sin^4 x\cos ^2 x dx=I_6-I_4$ so $I_6=-\cos x \sin^5 x + 5I_6-5I_4\Rightarrow I_6=\frac{\cos x \sin^5x + 5I_4}{4}$.

But, $I_4=$ $\int(\sin^2 x)^2dx=\int\left(\frac{1-\cos 2x}{2}\right)^2dx=\frac{1}{4}\left\{\int dx-\int 2\cos 2xdx+\int\cos^22xdx\right\}$ and $\int 2\cos 2x = \sin 2x +C$, $\int\cos^22xdx=\int\frac{1+\cos4x}{2}dx=\frac{x}{2 }+\frac{\sin 4x}{8}+K$.

Putting everything back in order, you'll get your primitive.

You may also, instead of calculating $I_4$ like I did, re-integrate by parts and you'll have to calculate $I_2$ which shouldn't be a problem using that $\sin^2 x = \frac{1-\cos 2x}{2}$