integral of sin^3 cos x dx

**Craka** · June 4th 2008, 04:22 PM

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

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

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

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

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

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

The answer is:

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

**Mathstud28** · June 4th 2008, 04:26 PM

Originally Posted by Craka

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

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

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

The answer is:

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

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

back subbing we get

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

**CaptainBlack** · June 4th 2008, 08:16 PM

Originally Posted by Craka

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

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

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

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

RonL

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