1. ## \int{\frac{x^2}{\sqrt{4x-x^2}}}dx

Having trouble with this one.

$\displaystyle \int{\frac{x^2}{\sqrt{4x-x^2}}}dx$

2. Originally Posted by th%$&873 Having trouble with this one.$\displaystyle \int{\frac{x^2}{\sqrt{4x-x^2}}}dx$Note that$\displaystyle 4x - x^2 = 4 - (x + 2)^2$. Make the substitution$\displaystyle x + 2 = 2 \sin t$. You might want to check the details (as I rushed this) - I get something like$\displaystyle 8 \int (\sin t - 1)^2 \, dt$. Expand, integrate term-by-term and finally substitute back that$\displaystyle x + 2 = 2 \sin t$(tedious, I know, but not too hard). Ask for more help if you get stuck at any point. Edit: There is a small mistake - you should spot it easily. A small detail that doesn't change the general game plan ..... 3. Originally Posted by th%$&873
Having trouble with this one.

$\displaystyle \int{\frac{x^2}{\sqrt{4x-x^2}}}dx$
$\displaystyle \int{\frac{x^2}{\sqrt{4x-x^2}}}\:dx = \int \frac{x^2}{\sqrt{4-(x-2)^2}}\:dx$

Let $\displaystyle x-2=2\sin u$ then,

$\displaystyle \int \frac{x^2}{\sqrt{4-(x-2)^2}}dx=4\int \sin^2 u+2\sin u+1\:du$, now you integrate each term

now you should get, $\displaystyle 4\int \sin^2 u+2\sin u+1\:du=6u-2\sin u\cos u-8\cos u$

$\displaystyle \int{\frac{x^2}{\sqrt{4x-x^2}}}\;dx=6\arcsin\left(\frac{x-2}{2}\right)-4\sqrt{4x-x^2}-\frac{(x-2)\sqrt{4x-x^2}}{2}+C$

4. Originally Posted by polymerase
$\displaystyle \int{\frac{x^2}{\sqrt{4x-x^2}}}\:dx = \int \frac{x^2}{\sqrt{4-(x-2)^2}}\:dx$

Let $\displaystyle x-2=2\sin\:u$ then,

$\displaystyle \int \frac{x^2}{\sqrt{4-(x-2)^2}}dx=4\int \sin^2\:u+2\sin\:u+1\:du$, now you integrate each term

now you should get, $\displaystyle 4\int \sin^2\:u+2\sin\:u+1\:du=6u-2\sin u\cos u-8\cos\:u$

$\displaystyle \int{\frac{x^2}{\sqrt{4x-x^2}}}\;dx=6arcsin\left(\frac{x-2}{2}\right)-4\sqrt{4x-x^2}-\frac{(x-2)(\sqrt{4x-x^2})}{2}+C$
5. I don't really understand either of your methods (you combined two steps?), but I used u-substitution then trig-substitution to get $\displaystyle 6\theta-8\cos\:\theta-\sin 2\theta$.
It appears that polymerase used the double-angle formula to get $\displaystyle -2\sin u\cos u$ instead of $\displaystyle -\sin 2 u$? I just wasn't sure what to do with that last term. I also made a small mistake with the coefficient of the first term, but I caught that the second time around.
6. Originally Posted by th%$&873 I don't really understand either of your methods (you combined two steps?), but I used u-substitution then trig-substitution to get$\displaystyle 6\theta-8\cos\:\theta-\sin 2\theta$. It appears that polymerase used the double-angle formula to get$\displaystyle -2\sin u\cos u$instead of$\displaystyle -\sin 2 u$? I just wasn't sure what to do with that last term. I also made a small mistake with the coefficient of the first term, but I caught that the second time around. You're right, I did use the double angle formula, the reason is because I needed it when I'm re-substituting and expressing everything in terms of$\displaystyle x\$. Thats all.