Question

Intergrate this problem please?

Answer 1

\[\int x\sqrt\frac{1-x^2}{1+x^2}~dx\]
Try a substitution, \(u=1+x^2\), then \(du=2x~dx\).
\[\frac{1}{2}\int \sqrt\frac{1-(u-1)}{u}~du\]

Answer 2

then it equals to 1/2 int ( sqrt ( 2-u / u )) ?

Answer 3

then do I separate these two?

Answer 4

Yes to the \((2-u)/u\). I'm wondering what we can do next... Maybe multiplying by \(\sqrt{\dfrac{u}{u}}\) might help?
\[\frac{1}{2}\int \sqrt\frac{2-u}{u}\frac{\sqrt u}{\sqrt u}~du=\frac{1}{2}\int \frac{\sqrt{2u-u^2}}{u}~du\]

Answer 5

maybe multiply top and bottom with sqrt(2-u)?

Answer 6

I'm a bit confused on this haha xD

Answer 7

You might be able to do a trig sub from here. Complete the square first:
\[2u-u^2=1-1+2u-u^2=1-(1-u)^2\]
So then a trig sub, \(1-u=\sin t\), so \(-du=\cos t~dt\):
\[\frac{1}{2}\int\frac{\sqrt{1-(1-u)^2}}{u}~du=-\frac{1}{2}\int \frac{\sqrt{1-\sin^2t}}{1-\sin t}\cos t~dt\]
Simplify:
\[-\frac{1}{2}\int \frac{\sqrt{\cos^2t}}{1-\sin t}\cos t~dt\\
-\frac{1}{2}\int \frac{\cos t}{1-\sin t}\cos t~dt\\
-\frac{1}{2}\int \frac{\cos^2 t}{1-\sin t}~dt\]
This seems a bit roundabout, but try this next:
\[-\frac{1}{2}\int \frac{\cos^2 t}{1-\sin t}\frac{1+\sin t}{1+\sin t}~dt\]
So we can simplify some more:
\[-\frac{1}{2}\int \frac{\cos^2 t(1+\sin t)}{1-\sin^2 t}~dt\\
-\frac{1}{2}\int \frac{\cos^2 t(1+\sin t)}{\cos^2 t}~dt\\
-\frac{1}{2}\int (1+\sin t)~dt\]
That should do it I think... Hope I didn't make a mistake.

Answer 8

wait how did you go from cos^2t (1+sint)/cos^2 t
to 1+ sint?
and OMG THIS IS AMAZING x.x

Answer 9

Don't the squared cosines cancel?

Answer 10

OH, stupid me xD Thank you!! so much! You're amazing :)

Answer 11

You're welcome!

Answer 12

@SithsAndGiggles there's one about log that I don't understand,... Last one, I promise! (these are challenge problems my tutor gave me, so it's really hard). Can you please help?

Answer 13

I'll make a new post