1. ## indefinite integrals

Hi Everyone,

Just to make sure I did it right, everytime I'm ok a differente rule arise

Here is my Question and answer

Find the indefinite integrals of the following functions.

(i)h(u) = cos2(3u)

∫ cos2(3u) du

∫ 1/3 sin2 (3u) = 1/3 sin2 + 3u + c

Thanks again for helping me

2. Originally Posted by valerie-petit
Hi Everyone,

Just to make sure I did it right, everytime I'm ok a differente rule arise

Here is my Question and answer

Find the indefinite integrals of the following functions.

(i)h(u) = cos2(3u)

∫ cos2(3u) du

∫ 1/3 sin2 (3u) = 1/3 sin2 + 3u + c

Thanks again for helping me
Sigh...

I assume this is

$\int\cos^2(3u)du$

and not

$\cos(3(2u))du$

FYI holding the alt key, while holding it press 0178 and the realease it, it will create ²...0179 gives ³.

But anyways

$\cos^2(3u)=\frac{1+\cos(6u)}{2}$

So we have

$\frac{1}{2}\int\bigg[1+\cos(6u\bigg]du=\frac{1}{2}\bigg[u+\frac{1}{6}\sin(6u)\bigg]+C$

3. Thanks again Mathstud28 for your help, you right it is cos²(3u)du