1. ## Integration

hi I'm having trouble integrating this

$\displaystyle \int sin^3\theta\cdot 3sin\theta cos\theta$

I don't really have a method for integrating trig it's normally pretty simple to work out but i don't know where to start with this one :/

2. Originally Posted by djmccabie
hi I'm having trouble integrating this

$\displaystyle \int sin^3\theta\cdot 3sin\theta cos\theta$

I don't really have a method for integrating trig it's normally pretty simple to work out but i don't know where to start with this one :/
$\displaystyle 3\int \sin^4{\theta} \cos{\theta} \, d\theta$

integration by substitution ...

let $\displaystyle u = \sin{\theta}$

3. ah thanks! how to you know what to let u equal?

4. Hello,
Originally Posted by djmccabie
ah thanks! how to you know what to let u equal?
By noting that cos is the derivative of sin.
So you'll have something in the form $\displaystyle u^4*u'$, and then, it's just recognizing known antiderivatives
And they're more easily spotted by making a substitution

5. yes i realised while i was answering the question

6. ok so this question actually has limits $\displaystyle \int_{0}^{\frac{\pi}{2}}$

now i have the mark scheme for this question which shows that the limits stay the same. But i thought when you did integration by substitution you had to manipulate the limits :/

7. Originally Posted by djmccabie
ok so this question actually has limits $\displaystyle \int_{0}^{\frac{\pi}{2}}$

now i have the mark scheme for this question which shows that the limits stay the same. But i thought when you did integration by substitution you had to manipulate the limits :/
You have two possibilities, while making a substitution :

- you substitute, and you change the boundaries
- you substitute, and you don't change the boundaries. In which case, you have to «back substitute», that is : after you got an antiderivative in terms of the new variable, substitute it with the old variable, and keep the limits.

But I think the first one is better, because for the second one, it would be more correct to write :

$\displaystyle \int_{\theta=0}^{\theta=\pi/2} u^4 ~du$, so that you know that it's the boundaries for theta.
While for the first situation, you can just say $\displaystyle \int_0^1 u^4 ~du$

Got it ?

8. so why limits $\displaystyle \int_{0}^{1}$ ??

i was under the impression you had to make theta = 0 to work out the bottom limit, then theta = pi/2 for the top limit :/

9. Originally Posted by djmccabie
so why limits $\displaystyle \int_{0}^{1}$ ??

i was under the impression you had to make theta = 0 to work out the bottom limit, then theta = pi/2 for the top limit :/
$\displaystyle \theta=0 \Rightarrow u=\sin\theta=0$
$\displaystyle \theta=\pi/2 \Rightarrow u=\sin\theta=1$

Be careful of the order of the boundaries.
In the case where you let $\displaystyle y=\cos\theta$, you would have $\displaystyle \int_1^0 =-\int_0^1$
But that's another story ^^

10. Thanks!

would i integrate this by substitution?

$\displaystyle \frac{1}{2} \int_{0}^{\frac{\pi}{2}} sin^22\theta d\theta$

or change into compound angle i.e.

$\displaystyle sin2\theta$ = $\displaystyle 2sin\theta cos\theta$ and then square it?

or does the square prevent this?

11. $\displaystyle \sin ^2 2\theta = 1 - \cos 4\theta$