help with trig integral: (3/4) int [ cos^2(2x) ] dx

[mahaney03]

Hi, I'm trying to figure out how my teacher got an answer for a problem
The integral is \(\displaystyle \L\\\frac{3}{4}\int{cos^{2}(2x)}dx\)

also i'm not exactly sure but i was wondering if my logic was right. I've been thinking the int of cos 2x is 1/2 cos 2x and wondering if im right or if my mind is confuzzled or something.

also another problem i was having trouble with was

\(\displaystyle \L\\6\int\frac{1}{cos^{2}(3x)}dx=6\int{sec^{2}(3x)}dx\) supposedly the answer is suppose to be 2 tan (3x) + C but i don't understand how he came to that answer

if anyone can help out it'd be greatly appreciated

edit: I'm also having trouble with integrating \(\displaystyle \L\\\int{x^{2}cos(3x)}dx\)
I'm trying to use integration by parts. There is supposed to be an integral from the tablei can use to solve it easily but i couldnt find it or rather, i couldnt figure out the right formula to use

 

[galactus]

The integral is \(\displaystyle \L\\\frac{3}{4}\int{cos^{2}(2x)}dx\)

Use the identity \(\displaystyle \L\\cos^{2}(2x)=\frac{1}{2}+\frac{1}{2}cos(4x)\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(\displaystyle \L\\6\int\frac{1}{cos^{2}(3x)}dx=6\int{sec^{2}(3x)}dx\) supposedly the answer is suppose to be 2 tan (3x) + C but i don't understand how he came to that answer

\(\displaystyle \L\\\int{sec^{2}(3x)}dx=\int[\underbrace{1+tan^{2}(3x)}_{\text{=sec^2(3x)}}]dx=\frac{tan(3x)}{3}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I'm also having trouble with integrating \(\displaystyle \L\\\int{x^{2}cos(3x)}dx\)

\(\displaystyle \L\\\int{x^{2}cos(3x)}dx\)

Let \(\displaystyle \L\\u=x^{2}, \;\ dv=cos(3x)dx, \;\ du=2xdx, \;\ v=\frac{1}{3}sin(3x)\)

Parts:

\(\displaystyle \L\\\frac{x^{2}}{3}sin(3x)-\frac{2}{3}\int{xsin(3x)}dx\)

parts again:

Let \(\displaystyle \L\\u=x, \;\ dv=sin(3x)dx, \;\ du=dx, \;\ v=\frac{-cos(3x)}{3}\)

\(\displaystyle \L\\\frac{x^{2}}{3}sin(3x)+\frac{2}{9}xcos(3x)-\frac{2}{27}sin(3x)\)