
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...
$\displaystyle
fsin^4xcos^2x dx
$
I've gotten this far:
$\displaystyle fsin^4x(1sin^2x)dx$
$\displaystyle fsin^4x dx  fsin^6x dx$
I am not sure how to proceed from this point. Any help is appreciated!

Let $\displaystyle I_n=\int\sin^n xdx$, then using partial integration $\displaystyle I_6=\int(\cos x)'\sin^5xdx=\cos x\sin ^5 x + 5\int \sin^4 x\cos ^2 x dx$ but $\displaystyle \int \sin^4 x\cos ^2 x dx=I_6I_4$ so $\displaystyle I_6=\cos x \sin^5 x + 5I_65I_4\Rightarrow I_6=\frac{\cos x \sin^5x + 5I_4}{4}$.
But, $\displaystyle I_4=$$\displaystyle \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 $\displaystyle \int 2\cos 2x = \sin 2x +C$, $\displaystyle \int\cos^22xdx=\int\frac{1+\cos4x}{2}dx=\frac{x}{2 }+\frac{\sin 4x}{8}+K$.
Putting everything back in order, you'll get your primitive.
You may also, instead of calculating $\displaystyle I_4$ like I did, reintegrate by parts and you'll have to calculate $\displaystyle I_2$ which shouldn't be a problem using that $\displaystyle \sin^2 x = \frac{1\cos 2x}{2}$