Hi
how to integrate this:
integral ((sin(x))^2/ sec(x)) dx)
what i did was:
1/sec(x) = cos(x)
let u = sin(x)
du = cos(x)
u^2 = sin(x)^2
integral of u^2 du
= 1/3u^3
is that right?
Looks good; remember that you've not finished, though! When you substitute, you have to remember to revert back to the form of the question. The original problem is in terms of x, not u, so your answer needs to be in terms of x.
Anyway, here's how I would solve this:
$\displaystyle \displaystyle\int\frac{sin^2(x)}{sec(x)}dx$
$\displaystyle =\displaystyle\int Sin^2(x)Cos(x)dx$
Are you aware that $\displaystyle \displaystyle\int f'(x)f(x)^n = \frac{f(x)^{n+1}}{n+1} + C$? This is derived from using the chain rule in reverse.
If you're not, then stick with substitution for now, but this is a lovely place to apply this rule because:
$\displaystyle f'(x)=Cos(x)$
$\displaystyle f(x)=Sin(x)$
$\displaystyle n=2$
So you literally just have to substitute into the formula with no difficult work required at all.
$\displaystyle \displaystyle\int Sin^2(x)Cos(x)dx = \frac{Sin^3(x)}{3} + C$
Hmm for some reason my assignment shows the solution as wrong.
Maybe because it's (sin(x))^2?
I thought that would be the same thing as sin^2(x) however so I'm not sure why its complaining.
Edit: The integral is part of a bigger question as shown above. I'm supposed to solve that differential equation. The left side is already done for me and i had to do the right. I'm pretty sure thats how I had to solve it?
Separating variables gives
$\displaystyle \displaystyle \int ye^{-y}dy=\int \sin^2(\theta)\cos(\theta)d\theta$
Using parts on the left and what you did before gives
$\displaystyle -ye^{-y}-e^{-y}=\frac{\sin^3(\theta)}{3}+C$ or
$\displaystyle -(y+1)e^{-y}=\frac{\sin^3(\theta)}{3}+C$