Trig Substitution

qbkr21 · #1 $Old$ October 1st, 2007, 06:18 PM

Thanks!!

Krizalid · #2 $Old$ October 1st, 2007, 08:43 PM

$\int\sin^4x\cdot\cos^2x\,dx=\int\sin^4x\,dx-\int\sin^6x\,dx$

From here you can use the identity $\cos2x=\cos^2x-\sin^2x$

It's messy, I hate this integral.

red_dog · #3 $Old$ October 2nd, 2007, 10:35 AM

$\displaystyle I_n=\int\sin^nxdx=\int\sin^{n-1}x\sin xdx=\int\sin^{n-1}x(-\cos x)'dx=$
$\displaystyle=-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}\cos^2dx=$
$\displaystyle=-\sin^{n-1}x\cos x+(n-1)\int(\sin^{n-2}x-\sin^nx)dx=$
$\displaystyle=-\sin^{n-1}x\cos x+(n-1)I_{n-2}-(n-1)I_n$

Then $I_n=-\frac{1}{n}\sin^{n-1}x\cos x+\frac{n-1}{n}I_{n-2}$

Krizalid · #4 $Old$ October 2nd, 2007, 10:36 AM

Yeah, but we actually don't know if qbkr21 may use reduction formulas.