Question

integral of sin(3x) cos^2(3x) dx

Answer 1

The answer is:
\[-\frac{1}{9} \cos ^3(3 x)
\]
Make the change of variable u=cos(3x) du = - 3 sin(3 x) dx and you can make the final step.

Answer 2

@ncorrea92 did you understand how to solve the problem?

Answer 3

Not really

Answer 4

substitute\[u=\cos \left( 3x \right)\]\[du=-3\sin \left( 3x \right)dx\]\[-\frac{1}{3}du=\sin \left( 3x \right)dx\]

Answer 5

Thanks i got it

Answer 6

Okay:) we have
\[\int \sin (3x) \cos^2 (3x) dx\]
We notice that there is sin 3x and cos 3x , so we can use substitution here, either sin 3x or cos 3x
if we substitute sin 3x it won't help much
Let's substitute
\[\cos 3x = u\]
differentiate this with respect to x
\[-3 \sin 3x=\frac{du}{dx}\]
so
\[-3\ \sin 3x\ dx= du\]
or
\[\sin 3x\ dx =-\frac{du}{3}\]
Let's substitute this in the integral
\[\int \sin 3x\ \cos^2 3x\ dx\]
let's substitute now
\[\int -u^2\ \frac{du}{3}\]
so we get now
\[-\frac{u^3}{3 \times 3} +c\]
\[-\frac{u^3}{9}+c\]
u= cos 3x
so we get
\[-\frac{\cos^3 3x}{9}+c\]
c= is a constant