# Thread: Help with an integral

1. ## Help with an integral

$\displaystyle \int sin^2(x) cos^3(x) dx$
My answer came out to be?
$\displaystyle sin^5x/5 - sin^3x/3$

2. ## correct!

Originally Posted by xterminal01
$\displaystyle \int sin^2(x) cos^3(x) dx$
My answer came out to be?
$\displaystyle sin^5x/5 - sin^3x/3$
your answer seems to be correct. there will be 2 sins one with power 4 and the other with power 2 and a cos which will go with the dx once you get dt if t=sinx.but u must not forget the constant c!

3. Originally Posted by xterminal01
$\displaystyle \int sin^2(x) cos^3(x) dx$
My answer came out to be?
$\displaystyle sin^5x/5 - sin^3x/3$
Write the integral as $\displaystyle \int sin^2(x) cos^2(x) cos(x)dx= \int sin^2(x)(1- sin^2(x)) cos(x)dx$$\displaystyle = \int (sin^2(x)- sin^4(x)) cos(x)dx Now let u= sin(x) so that du= cos(x)dx and the integral becomes \displaystyle \int (u^2- u^4)du$$\displaystyle \frac{u^3}{3}- \frac{u^5}{5}+ C= \frac{sin^3(x)}{3}- \frac{sin^5(x)}{5}$, the negative of your answer.

It is always better, in asking for a check, to show your work.