1. ## Integrating trig functions

Hi, I'm doing some questions on the mean value theorem and while I understand the theory in my set of answers they tend to jump from integrating compex trig functions to just giving the answer so if someone could explain how to go from

$1$/ $2\pi$ $\int_{0}^{2\pi}$ $\cos^2$ $\theta$ $\sin$ $\theta$ $d$ $\theta$

-(cos^3 $\theta$\6) between $0$ and $2\pi$ which obviously equals 0. If someone could give me a step by step explanation of how to get from the initial integral to the answer I'd be very greatful. I've tried plugging in quite a few different trig identities but I just get into a mess!!

Cheers!

P.S. sorry I gave up with latex.. too frustrating to get it to give the right output but hopefully post still makes sense!

2. Originally Posted by shmounal
Hi, I'm doing some questions on the mean value theorem and while I understand the theory in my set of answers they tend to jump from integrating compex trig functions to just giving the answer so if someone could explain how to go from

$1$/ $2\pi$ $\int_{0}^{2\pi}$ $\cos^2$ $\theta$ $\sin$ $\theta$ $d$ $\theta$

-(cos^3 $\theta$\6) between $0$ and $2\pi$ which obviously equals 0. If someone could give me a step by step explanation of how to get from the initial integral to the answer I'd be very greatful. I've tried plugging in quite a few different trig identities but I just get into a mess!!

Cheers!

P.S. sorry I gave up with latex.. too frustrating to get it to give the right output but hopefully post still makes sense!
Use the substitution $u = cos(x) \implies du = -sin(x)~dx$.

-Dan

3. \begin{aligned} & I = \frac{1}{2\pi}\int_{0}^{2\pi}\cos^2{\theta}\sin{ \theta}\;{d\theta} = \frac{1}{2\pi}\int_{0}^{2\pi}\cos^2(2\pi-\theta)\sin(2\pi-\theta)\;{d\theta} = -\frac{1}{2\pi}\int_{0}^{2\pi}\cos^2{\theta}\sin{ \theta}\;{d\theta}. \\& \therefore ~~ I+I = \frac{1}{2\pi}\int_{0}^{2\pi}\cos^2{\theta}\sin{ \theta}\;{d\theta}-\frac{1}{2\pi}\int_{0}^{2\pi}\cos^2{\theta}\sin{ \theta}\;{d\theta} = 0 ~~ \Rightarrow 2I = 0 \Rightarrow I = 0. \end{aligned}

4. @Dan

oh yeah, as I thought easy when you know how. Is there any clever way to notice when to sub in or when to juggle trig identies? I'm willing to bet you'll say its just a matter of experience but worth an ask as this is by far the biggest problem I have with my Maths. I always seem to understand the concept but when it comes to actually integrating or differentiating functions I get in a big jumble!!

@Coffee Machine, thanks thats pretty nifty too!

Cheers

5. Originally Posted by shmounal
@Dan

oh yeah, as I thought easy when you know how. Is there any clever way to notice when to sub in or when to juggle trig identies? I'm willing to bet you'll say its just a matter of experience but worth an ask as this is by far the biggest problem I have with my Maths. I always seem to understand the concept but when it comes to actually integrating or differentiating functions I get in a big jumble!!
Yeah, pretty much experience. However this is part of the usual substitution method: if you have
$\int f(x)~f'(x)~dx$

then try the substituion u = f(x). You get more comfortable with them the more you work them out. And hey, people still have to point some of them out to me. So don't worry about it. When you see this one next time, you'll know what to do.

-Dan

6. Originally Posted by shmounal
oh yeah, as I thought easy when you know how. Is there any clever way to notice when to sub in or when to juggle trig identies? I'm willing to bet you'll say its just a matter of experience but worth an ask as this is by far the biggest problem I have with my Maths. I always seem to understand the concept but when it comes to actually integrating or differentiating functions I get in a big jumble!
I think one of the most effective ways of learning integration techniques is toying with problems. Make a substitution and see if it works; if it it doesn't work, try to see why. If it works, try to find another method, or substitution. That way, you will develop a very good intuition as to which approach is more appropriate, as well as find many interesting techniques. For example, you could have 'juggled' through trigonometric identities and found the integral that way, but Dan's substitution is far better, as you would find. So toying with this particular integral itself would aid your intuition as to which approach is more appropriate for a particular integral. In post two, we had put θ ↦ 2π-θ, which is made handy by the identity:

$\displaystyle \int_{a}^{b}f(x)\;{dx} = \int_{a}^{b}f(a+b-x)\;{dx}$.

(See here for more about it). Our integral was not terribly interesting/hard, but toying with, as I said earlier, could lead you to discover many interesting techniques which, while being perhaps overkill for this integral, are nonetheless useful for harder integrals. One such technique introduces a related integral, as illustrated below:

\begin{aligned} & I := \frac{1}{2\pi}\int_{0}^{2\pi}\cos^2{\theta} \sin{\theta}\;{d \theta}, ~~ J := \frac{1}{2\pi}\int_{0}^{2\pi}\sin^3{\theta} \;{d \theta} \\& \begin{aligned}\Rightarrow I+J & = \frac{1}{2\pi}\int_{0}^{2\pi}\cos^2{\theta} \sin{\theta}+\sin^3{\theta}\;{d\theta} \\& = \frac{1}{2\pi}\int_{0}^{2\pi}\sin{\theta}\left( \cos^2{\theta}+\sin^2{\theta}\right)\;{d\theta} \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}\sin{\theta}\;{d \theta} \\& = 0. \end{aligned} \\& \begin{aligned}\Rightarrow I-J & = \frac{1}{2\pi}\int_{0}^{2\pi}\cos^2{\theta} \sin{\theta}-\sin^3{\theta}\;{d\theta} \\& = \frac{1}{2\pi}\int_{0}^{2\pi}\sin{\theta}\left( \cos^2{\theta}-\sin^2{\theta}\right)\;{d\theta} \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}\sin{\theta}\cos{2 \theta}\;{d \theta}\end{aligned} \end{aligned}

$\therefore ~~ (1+J)+(I-J) = 0+\frac{1}{2\pi}\int_{0}^{2\pi}\sin{ \theta}\cos{2\theta}\;{d \theta} \Rightarrow 2I = \frac{1}{2\pi}\int_{0}^{2\pi}\sin{\theta}\cos{2 \theta} \;{d \theta}.$

Again, define the (new) integral with a related integral:

\begin{aligned} & 2I := \frac{1}{2\pi}\int_{0}^{2\pi}\sin{\theta}\cos{2 \theta}\;{d\theta}, ~~ 2K := \frac{1}{2\pi}\int_{0}^{2\pi}\cos{\theta}\sin{2 \theta}\;{d \theta} \\& \begin{aligned}\Rightarrow 2I+2K & = \frac{1}{2\pi}\int_{0}^{2\pi}\sin{\theta}\cos{2 \theta}+\cos{\theta}\sin{2\theta}\;{d\theta} \\& = \frac{1}{2\pi}\int_{0}^{2\pi}\sin(\theta+2 \theta) \;{d \theta} \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}\sin{3\theta}\;{d \theta} \\& = 0. \end{aligned} \\& \begin{aligned}\Rightarrow 2I-2K & = \frac{1}{2\pi}\int_{0}^{2\pi}\sin{ \theta}\cos{2 \theta}-\cos{\theta}\sin{2 \theta}\;{d \theta} \\& = \frac{1}{2\pi}\int_{0}^{2\pi}\sin(\theta-2 \theta)\;{d\theta} \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}\sin(- \theta)\;{d \theta} \\& = -\frac{1}{2\pi}\int_{0}^{2\pi}\sin{ \theta}\;{d \theta} \\& = 0.\end{aligned} \end{aligned}

$\therefore 2I+2K+2I-2K = 0 \Rightarrow 4I = 0 \Rightarrow I = 0.$

Of course, we didn't have to go to all that trouble, but I wanted to stress the learning benefits of toying with problems. As a side note, I hate toying with geometry problems; whenever I'm confronted a problem that goes along the lines of 'let ABC be a triangle with such and such properties...' I have a thought interruption that says 'nah, it's impossible'! Hopefully university geometry won't put me off as much!

7. Wow, thats awesome thankyou. I started a uni maths degree after having about 5 years out from school so have found remembering all these techniques I used to take for granted really tricky and despite asking tutors for help or advice I've just been greeted with shrugs! Hopefully I can kick on from here. Thanks again!!