I am trying to do this:
integral of x^2/((4-x^2)^(5/2))
I set x = 2sin(u)
dx = 2cos(u)du
I am stuck at this point now:
integral of (2sin^2(u) * 2cos(u)du)/32cos^5(u)
some help plz. Thanks.
Hello, Dave!
I am trying to do this: .$\displaystyle \int\frac{x^2}{(4-x^2)^{\frac{5}{2}}}\,dx$
I set: .$\displaystyle x \,= \,2\sin\theta\quad\Rightarrow\quad dx \,=\,2\cos\theta\,d\theta$ . . . . Good!
I am stuck at this point now: .$\displaystyle \int\frac{2\sin^2\!\theta\cdot2\cos\theta\,d\theta }{32\cos^5\!\theta} $
We have: .$\displaystyle \frac{1}{8}\int\frac{\sin^2\!\theta}{\cos^4\!\thet a}\,d\theta \;= \;\frac{1}{8}\int\frac{\sin^2\!\theta}{\cos^2\!\th eta}\cdot\frac{1}{\cos^2\!\theta}\,d\theta\;=\;\fr ac{1}{8}\int\tan^2\!\theta\cdot\sec^2\!\theta\,d\t heta$
Then let: .$\displaystyle u \,=\,\tan\theta\quad\Rightarrow\quad du \,=\,\sec^2\!\theta\,d\theta$
Got it?
A minor correction:
$\displaystyle \int \frac{x^2}{(4 - x^2)^{5/2}} dx = \int \frac{4 sin^2(\theta) \cdot 2 cos( \theta) d \theta }{32cos^5(\theta)}$
You skipped a factor of 2. Then follow Soroban's suggestions.
I've seen at least three of these examples from you now and I'm getting the impression you are having some troubles with the actual substitution. My advice is for you to slow down and double check your work after you put the substitution in the integral, because you have a tendency to miss multiplicative factors.
-Dan