Hey guys,
Can you please show me a step by step integration for
Find the solution for the differential equation :
3Cos^2(x) , y= Pi , x = Pi/2
Thank you !
ummm, sort of. You do have to use double angle formulae if that's what you mean.
Think of this:
$\displaystyle \int cos^4 \theta \ d \theta=\int (cos^2 \theta)^2 \ d \theta$
We already know that (i'll show how Mr F gets his answer since you've never seen it before):
$\displaystyle cos^2 \theta +sin^2 \theta=1$ and $\displaystyle cos^2 \theta - sin^2 \theta= cos 2\theta$.
We require an expression for $\displaystyle cos^2 \theta$.
$\displaystyle cos^2 \theta +sin^2 \theta=1 \Rightarrow \ sin^2 \theta = 1-cos^2 \theta$ from the first identity.
Putting this into the second identity gives:
$\displaystyle 2cos^2 \theta -1=cos 2 \theta \Rightarrow \ cos^2 \theta = \frac{cos 2 \theta +1}{2}$.
Going back to our original integral:
$\displaystyle \int cos^4 \theta \ d \theta=\int (cos^2 \theta)^2 \ d \theta = \ \int \left( \frac{cos 2 \theta +1}{2} \right)^2 d \theta$$\displaystyle =\frac{1}{4} \int (cos 2 \theta +1)^2 d \theta=\frac{1}{4} \int cos^2 2 \theta +2 cos 2 \theta +1 d \theta$.
The big problem now is the $\displaystyle cos^2 2 \theta$.
Looking at the integral you posted and what I have done here, can you finish this integral off?
(Hint: $\displaystyle cos^2 (2 \theta)+sin^2 (2 \theta)=1$ and $\displaystyle cos^2 (2 \theta)- sin^2 (2 \theta)=cos(4 \theta)$).