How do you integrate (cos 3x)(cos x) with respect to x?

Answer 1

First use defactorisation formula to simplify.

2cosAcosB = cos(A+B) + cos(A-B)
thus (cos 3x)(cos x)=(1/2)*[cos(3+1)x + cos(3-1)x]

cos 3x)(cos x) = (1/2)*[cos4x + cos2x]

Now Integrate by using formula for cos(a*x) :

I = (1/2)*{(sin4x)/4 + (sin2x)/2} + c
I = (sin4x)/8 + (sin2x)/4 + c
where c is the constant of integration.

Answer 2

Just express it as a sum of cos's

2Cos(x)cos(y) = Cos(x+y) + Cos(x-y)

So
2cos(3x)cos(x) = cos(4x) +cos(2x)

Integral((cos 3x)(cos x)) =
Integral((cos(4x) +cos(2x)) / 2) =
sin(4x)/8 + sin(2x)/4 + C

Answer 3

u can write this in the form of 2(cos 3x)(cos x)/2
it doesnt amke any change like this.
now,we know
2Cos(x)cos(y) = Cos(x+y) + Cos(x-y)

2cos(3x)cos(x) = cos(4x) +cos(2x)

Integral((cos 3x)(cos x))= Integral((cos(4x) +cos(2x)) / 2)
= sin(4x)/8 + sin(2x)/4 + C