how to integrate 4cos^2(3x)

Question

ganeshie8 · Answer

rewrite $\cos^2x$ as $(\cos(2x) + 1) / 2$

shamim · Answer

u hv to use a trigonometric formula here
2cos^A=1+cos2A

anonymous · Answer

how to integrate 4cos6x+2 then?

ganeshie8 · Answer

$\large  \int 4\cos^2(3x)  ~dx$


$\large  \int 2\times  2\cos^2(3x)  ~dx$


$\large  \int 2( 1 + \cos  (2\times 3x)) ~dx$


$\large  \int 2 + 2\cos  (6x) ~dx$

ganeshie8 · Answer

you got that right ?

anonymous · Answer

Yeah I got 2+2cos6x. sorry for the mistake earlier ><

ganeshie8 · Answer

$\large  \int 2 + 2\cos  (6x) ~dx$


split the integral :


$\large  \int 2~ dx     +  \int 2\cos  (6x) ~dx$


$\large  2\int 1~ dx     +  2\int \cos  (6x) ~dx$

ganeshie8 · Answer

whats the integral :    $\int 1~ dx$
 ?

anonymous · Answer

x?

ganeshie8 · Answer

yes, 
wat about the integral : $\int \cos (x)  ~dx$
?

anonymous · Answer

sin x?

ganeshie8 · Answer

yes, but since we have cos(3x), we need to fix it a bit :

$\int \cos(3x)~dx  =   \frac{\sin(3x)}{3} + c$

ganeshie8 · Answer

So,

$\large  2\int 1~ dx     +  2\int \cos  (6x) ~dx$


evaluates to :


$\large  2x   +  2\frac{\sin(6x)}{6}  + C$

ganeshie8 · Answer

simplify if u can

anonymous · Answer

Omg, thank you so much!!!

ganeshie8 · Answer

u wlc :)