Find the integral of sinxcosxcos2x dx



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nathanshields's profile pic

Posted on (Answer #1)

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


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:


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`

Top Answer

embizze's profile pic

Posted on (Answer #2)

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`




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

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

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




`=sinxcosxcos2x` as required. **

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