# Thread: Indefinite integrals

1. ## Indefinite integrals

Compute

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

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

2. Originally Posted by liyi $\displaystyle \int\frac{\sin x\cos x}{1+2\sin^2x\cos^2x}\,dx$
Let $\displaystyle t=\sin^2 x$ then $\displaystyle t ' = 2\sin x \cos x$ and $\displaystyle \cos^2 x = 1-t^2$.

3. 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.

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