# Math Help - Integrals...please...help

1. ## Integrals...please...help

Hi,
Im new!I have been having trouble with this calculus stuff in general, but integrals have tripped me up wayyy more than derivatives and i need major help!

okay so here is the problem im stuck on

INTEGRATE: (1-sin23t) cos 3t dt

so do i substitute or do i do some other math thingy?

Sorry for my ignorance,

Thanks in Advanced

2. ## Re: Integrals...please...help

Originally Posted by IgnorantMathGuru
Hi,
Im new!I have been having trouble with this calculus stuff in general, but integrals have tripped me up wayyy more than derivatives and i need major help!

okay so here is the problem im stuck on

INTEGRATE: (1-sin23t) cos 3t dt

so do i substitute or do i do some other math thingy?

Sorry for my ignorance,

Thanks in Advanced
you should know another form of $1-\sin^2(3t)$ that would be convenient for this problem.

once you figure that out the integral should be simple

3. ## Re: Integrals...please...help

is the answer 1/3 sin 3t+c

because my answer key looks like its saying something different

4. ## Re: Integrals...please...help

Originally Posted by IgnorantMathGuru
is the answer 1/3 sin 3t+c

because my answer key looks like its saying something different
no that's not the answer

5. ## Re: Integrals...please...help

is this the answer:
1/3sin3t-1/9 sin^3t+c

if so....can... you help me get to that answer maybe?

6. ## Re: Integrals...please...help

Originally Posted by IgnorantMathGuru
is this the answer:
1/3sin3t-1/9 sin^3t+c

if so....can... you help me get to that answer maybe?
$\displaystyle{\int}\left(1-\sin^2(3t)\right)\cos(3t)~dt$

$u=\sin(3t)$

$du=3\cos(3t)dt$

$\displaystyle{\int}(1-u^2)~\dfrac {du} 3$

$\dfrac 1 3 \left(u-\dfrac{u^3}{3}\right)+C$

$\dfrac 1 3 \left(\sin(3t)-\dfrac{\sin^3(3t)}{3}\right)+C$