# Thread: integral of sin^3 cos x dx

1. ## integral of sin^3 cos x dx

Definately having some problems with these.
Question is evalueate the indefinate integral

$\displaystyle \int {\sin ^3 } x\_\cos x\_dx$

I get as far as this, I seem to be having issues with these.

$\displaystyle \int {\sin ^2 } x\_\sin x\_\cos x\_dx$

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

$\displaystyle u = \cos x$ $\displaystyle du = - \sin x$

$\displaystyle \frac{1}{4}\sin ^4 x + C$

2. Originally Posted by Craka
Definately having some problems with these.
Question is evalueate the indefinate integral

$\displaystyle \int {\sin ^3 } x\_\cos x\_dx$

I get as far as this, I seem to be having issues with these.
$\displaystyle \int {\sin ^2 } x\_\sin x\_\cos x\_dx \int {(1 - \cos ^2 } x)\sin x\cos x\_dx u = \cos x du = - \sin x$

$\displaystyle \frac{1}{4}\sin ^4 x + C$
No need for a use of trig identities either see that cos is the derivative of sin or do it the old fashioned way

Let $\displaystyle u=\sin(x)\Rightarrow{du=\cos(x)dx}$

giving $\displaystyle \int{u^3du}=\frac{u^4}{4}+C$

back subbing we get

$\displaystyle \frac{\sin^4(x)}{4}+C$

3. Originally Posted by Craka
Definately having some problems with these.
Question is evalueate the indefinate integral

$\displaystyle \int {\sin ^3 } x\_\cos x\_dx$
Since $\displaystyle \frac{d}{dx} \sin(x)=\cos(x)$ the integrand is of the form $\displaystyle [f(x)]^3f'(x)$, so the integral is $\displaystyle [f(x)]^4/4+C$, or:

$\displaystyle \int (\sin(x))^3 \cos(x)~dx = \frac{(\sin(x))^4}{4}+C$

RonL