Question

integrate sin3theta*sin5theta dtheta. I've done the problem a few times but keep coming up with the wrong answer.

Answer 1

What methods have you tried?
Hmm I think `by parts` will work very nicely.

Answer 2

I was using 1/2(cos(m-n)x-cos(m+n)x)

Answer 3

Hmm I don't know what that is D:

Answer 4

it's a set of three formulas we got in class. It was probably derived from integration by parts though originally

Answer 5

grrr you should have rotated it :) lol

Answer 6

haha, sorry

Answer 7

Hmm that's a nifty little formula.
So I guess that gives us,\[\large \int\limits \sin5\theta \sin3\theta\;d \theta \quad=\quad \frac{1}{2}\int\limits \cos2\theta-\cos8\theta\;d \theta\]right?

Answer 8

I did a -2theta for m-n

Answer 9

So you ended up with,\[\large \int\limits\limits \sin5\theta \sin3\theta\;d \theta \quad=\quad \frac{1}{2}\int\limits\limits \cos(-2\theta)-\cos8\theta\;d \theta\]

Answer 10

yes

Answer 11

Keep in mind that cosine is an `even function` so:\[\Large \cos(-2\theta) \quad=\quad \cos(2\theta)\]

Answer 12

ah, okay, cool

Answer 13

So what did you do from that step?
Did you do a `u-substition` ?

You'll want to try and get more comfortable with problems like this, so you don't have to bother with a u-sub.

On integrals like this, we simply divide by the coefficient on theta when we integrate.

Answer 14

Example:\[\Large \int\limits \cos(4\theta)\;d \theta \quad=\quad \frac{1}{4}\sin(4\theta)\]
Understand what I mean? :o
We just divide by the coefficient (the 4), instead of doing a fancy u-sub.
If that's too confusing you can go ahead with a u-sub though :)

Answer 15

that's what I was doing. The signs probably messed me up

Answer 16

yeah, definitely. I don't mess with u subs on those simple ones

Answer 17

\[\large \frac{1}{2}\int\limits\limits \cos2\theta-\cos8\theta\;d \theta \quad=\quad \frac{1}{2}\left(-\frac{1}{2}\sin(-2\theta)-\frac{1}{8}\sin8\theta\right)\]

Answer 18

So you end up with something like that?

Answer 19

Woops I meant to put cos(-2theta) in the original, to match what you've been working with.
Same result though.

Answer 20

that is what I've been getting

Answer 21

So why do you think your answer is wrong? :3
maybe it just looks a little different than the book.

Answer 22

:/

Answer 23

From here, you would want to remember that sine is an `odd function`:\[\Large \sin(-2\theta) \quad=\quad -\sin(2\theta)\]

Answer 24

So that's why that negative ends up disappearing on the first term.

Answer 25

We do our homework online and it keeps counting my answer wrong. Online systems are so picky though. Glad I'm doing it right though

Answer 26

Oh yah, that can be a bit tricky D:

Answer 27

A bit frustrating i mean*

Answer 28

Thank you for your help though. And btw, I've mastered that reduction formula now :P

Answer 29

Oh cool c: