How do I evaluate the indefinite integral $intsin^3(x)*cos^2(x)dx$ ?

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Answer 1 · 2014-09-21T23:59:49

The answer is $-(cos^3x)/3+(cos^5x)/5+C$ .

The trick with sinusoidal powers is to use identities so that you can have $sin x$ or $cos x$ with a power of 1 and use substitution.

In this case, it is easier to get $sin x$ to a power of 1 using $sin^2x=1-cos^2x$ .

$int sin^3x*cos^2x dx$
$=int sin x(1-cos^2x)cos^2x dx$
$=int sin x(cos^2x-cos^4x)dx$
$=int sin x cos^2xdx-int sin x cos^4x dx$

Now it is a matter of using substitution:

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

$int sin x cos^2xdx-int sin x cos^4x dx$
$=int -u^2 du+ int u^4 du$
$=-(u^3)/3+(u^5)/5+C$
$=-(cos^3x)/3+(cos^5x)/5+C$

How do I evaluate the indefinite integral intsin^3(x)*cos^2(x)dxintsin^3(x)*cos^2(x)dx ?