Compute

$\displaystyle \int\frac{2\sin^2x}{3-4\cos^2x}\,dx$

$\displaystyle \int\frac{\sin x\cos x}{1+2\sin^2x\cos^2x}\,dx$

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- Dec 15th 2007, 03:02 PMliyiIndefinite integrals
Compute

$\displaystyle \int\frac{2\sin^2x}{3-4\cos^2x}\,dx$

$\displaystyle \int\frac{\sin x\cos x}{1+2\sin^2x\cos^2x}\,dx$ - Dec 15th 2007, 03:23 PMThePerfectHacker
- Dec 15th 2007, 03:31 PMgalactus
You sure do have some tougher integration problems. May I ask what calculus class you are in?.

For #1:

$\displaystyle 2\int\frac{sin^{2}(x)}{3-4cos^{2}(x)}dx$

One way you could go about it is to let $\displaystyle x=tan^{-1}(x), \;\ dx=\frac{1}{u^{2}+1}du$

Make the subs and get:

$\displaystyle 2\int\frac{\frac{u^{2}}{u^{2}+1}}{3-4(\frac{1}{u^{2}+1})}\cdot\frac{1}{u^{2}+1}du$

This simplifies down to:

$\displaystyle \frac{1}{2}\int\frac{1}{3u^{2}-1}du+\frac{1}{2}\int\frac{1}{u^{2}+1}du$

Can you finish?. It's a matter of back to arctan and an ln.

You might also rewrite it as $\displaystyle \frac{cos(2x)-1}{2cos(2x)-1}$ and see what you can come up with. Different approaches.