Question

Integrate cos x cos 2x cos 3x w.r.t x

Answer 1

um guess u shud do it the long way

Answer 2

which one ?

Answer 3

u means the first expanding cos 2x and cos 3x and then using the product rule

Answer 4

wait cos2x= 2sin^2x -1

Answer 5

cos3x = cos(2x+x) expand it

Answer 6

nd put sinx=t

Answer 7

ok

Answer 8

Use \[\cos (A ) \cos (B )=\frac{1}{2} (\cos (A -B )+\cos (A+B))\]

Answer 9

\[\frac{1}{2}\int\limits \cos ^2(3 x) \, dx+\frac{1}{4}\int\limits (\cos (2 x)+\cos (4 x)) \, dx\]

Answer 10

i missed that :P

Answer 11

Ok got that.

but what about cos (2x) and cos (4x)

Answer 12

u=2x h=4x t=3x
du=2dx dh=4dx dt=3dx
\[\frac{1}{8}\int\limits \cos (u) \, du+\frac{1}{6}\int\limits \cos ^2(t) \, dt+\frac{1}{16}\int\limits \cos (h) \, dh\]

Answer 13

ok that would work for sure.

Thanks a lot for help

Answer 14

welcome :)