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Math Help - Indefinite integrals

  1. #1
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    Indefinite integrals

    Compute

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

    \int\frac{\sin x\cos x}{1+2\sin^2x\cos^2x}\,dx
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  2. #2
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    Quote Originally Posted by liyi View Post
    \int\frac{\sin x\cos x}{1+2\sin^2x\cos^2x}\,dx
    Let t=\sin^2 x then t ' = 2\sin x \cos x and \cos^2 x = 1-t^2.
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  3. #3
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    You sure do have some tougher integration problems. May I ask what calculus class you are in?.

    For #1:

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

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

    Make the subs and get:

    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:

    \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 \frac{cos(2x)-1}{2cos(2x)-1} and see what you can come up with. Different approaches.
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