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,

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,

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

is the answer 1/3 sin 3t+c

because my answer key looks like its saying something different

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

because my answer key looks like its saying something different

1/3sin3t-1/9 sin^3t+c

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

Originally Posted by IgnorantMathGuru
1/3sin3t-1/9 sin^3t+c

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

$u=\sin(3t)$

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

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

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

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