# Thread: Integration of Trigonometric Identities - Yay!

1. ## Integration of Trigonometric Identities - Yay!

Remind why I'm taking further maths next year >.<

Question time!

I'm currently doing, as you probably noticed in the title (:P) integration of trig identities. I've come across this lovely little question which asks me to:

Find the integral of cos2xcosx

(I have no idea how to stick the integral sign in a text page ^_^)

Would somebody be so kind as to explain to me how to do this?

Many thanks

2. Hello,

Note that cos(2x)=1-2sin²(x)

Hence cos(2x)cos(x)=cos(x)-2cos(x)sin²(x)

You know the primitive of cos(x) (at least i hope ^^)

For the second term, cos is the derivative of sin, so you have something of the form :

$k \ u'(x) \ u^2(x)$, which gives a primitive : $k \frac13 u^3(x)$

3. I'm just going to write down my chain of thought - could you please correct me if I go awry

cos(x) - 2cos(x)sin²(x)
cos(x) - 2cos(x)(1 - cos²(x))
cos(x) - 2cos(x) - 2cos^3 (x)
-cos(x) - 2cos^3 (x)

Is this right so far in terms of deriving etc. ?

4. Never mind, I'm going to call it a night and face the unparalleled wrath of an angry maths teacher

Thanks for the help guys ^_^ I have enough to make it look like I tried :P

X

5. Well, guess it's too late ~

cos(2x)=1-2sin²(x)

-> cos(2x)cos(x)=cos(x)-2cos(x)sin²(x)

$\int \cos(x) - 2 \cos(x) \sin^2(x) dx = \underbrace{\int \cos(x) dx}_{\sin(x)} - 2 \int \underbrace{\cos(x)}_{u'(x)} \underbrace{\sin^2(x)}_{u^2(x)} dx$

$= \sin(x) - 2*3 \int \frac13 u'(x) u^2(x) dx = \sin(x) - 6 [u^3(x)] = \sin(x)-6 \sin^3(x)$