# Find the integral of sinxcosxcos2x dx

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

Find `int (sinx cosx cos2x) dx`

(1) Let `u=cos2x` . Then `(du)=-2sin(2x)=-4sinxcosxdx`

(2) Substituting we get:

`int(sinxcosxcos(2x))dx=-1/4 int u du`

`=-1/4[u^2/2+C_1]`

`=-1/4[(cos^2(2x))/2+C_1]`

`=-1/8(cos^2(2x))+C`

`=-1/16 cos(4x)+C`

Thus the solution is `-1/16 cos(4x)+C`

** `d/(dx)[-1/16 cos(4x)+C]=-1/16(4)(-sin4x)`

`=1/4sin4x`

`=1/4[2sin2xcos2x]`

`=1/4[4sinxcosxcos2x]`

`=sinxcosxcos2x` as required. **

nathanshields | High School Teacher | (Level 1) Associate Educator

Posted on

You're going to have to use a couple of identities in this bad boy.  Start with this one to rewrite the integrand:

`cos(A)cos(B)=1/2(cos(A-B)+cos(A+B))`

So now you have `int sin(x)*1/2(cos(x)+cos(3x))dx`

(Remember, since cosine is an even function, cos(-x) = cos(x).) Now pop the 1/2 out front, distribute the sin(x), and break the beast in two:

`1/2 int sin(x)cos(x) dx + 1/2 int sin(x)cos(3x) dx`

Now it's time to bring in another identity:

`sin(A)cos(B)=1/2(sin(A-B)+sin(A+B))`

This identity turns "sin x cos x" into 1/2(sin(0)+sin(2x)) and
"sin x cos 3x" turns into 1/2(sin(-2x)+sin(4x).  Remember, since sine is an odd function, sin(-2x) = -sin(2x).  So now we have

`1/2 int 1/2 sin (2x) dx + 1/2 int 1/2(sin (4x) - sin (2x)) dx`

`=1/4 int sin(2x) dx + 1/4 int sin (4x) dx - 1/4 int sin(2x)dx`

Notice that the first and third terms cancel - cha-ching!  Now you take it from here:

`1/4 int sin(4x) dx`