1. ## integral

$\displaystyle \int sin^{4}x$

so what I did was trying to cut it up

$\displaystyle \int sin^{2}x(1-cos^{2}x)$

eventually I realized this method was wrong because I got a big fat zero

so any ideas???

2. Originally Posted by akhayoon
$\displaystyle \int sin^{4}x$

so what I did was trying to cut it up

$\displaystyle \int sin^{2}x(1-cos^{2}x)$

eventually I realized this method was wrong because I got a big fat zero

so any ideas???
$\displaystyle \int{sin^4(x)dx}=\int\bigg(\frac{1-cos(2x)}{2}\bigg)^2dx=\frac{1}{4}\int[1-2cos(2x)+cos^2(2x)]dx$$\displaystyle =\frac{1}{4}\int\bigg[1-2cos(2x)+\frac{1+cos(4x)}{2}\bigg]dx$...I think you can go from there

3. Originally Posted by akhayoon
$\displaystyle \int sin^{4}x$

so what I did was trying to cut it up

$\displaystyle \int sin^{2}x(1-cos^{2}x)$

eventually I realized this method was wrong because I got a big fat zero

so any ideas???
try the identity

$\displaystyle \sin^2(x)=\frac{1}{2}(1-\cos(2x))$

$\displaystyle \int (sin^2(x))^2dx=\int (\frac{1}{2}(1-\cos(2x)))^2dx$

$\displaystyle \frac{1}{4}\int 1-2\cos(2x)+\cos^2(2x)dx$

Now use the cos version of the above

$\displaystyle \cos^{2}(2x)=\frac{1}{2}(1+\cos(4x))$

Then just integrate.

Good luck.