- #1

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I'm currently learning integration by trig substitution. The problem is:

integral ( x^2/sqrt(16-x^2) ) dx.

I seem to be having some difficulty...this is what I have:

let x=4sinu

dx = 4cosu du

Thus, we have:

integral 16sin^2(u)*4cosu du/(4cos^2(u)) (the last part is from the trig identity sin^2x + cos^2x = 1)

But now what? I simplied that as much as I could and then used integration by parts, but I ended up with the wrong answer. The correct answer is 8arcsin(x/4) - 0.5x*sqrt(16-x^2). Have I made a mistake in what I've done so far? If not, where do I proceed?

Thanks,

LOS