Integral by Trig Substitution, Calc 2

Darkestsolrac
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Homework Statement


The definite integral of ∫(x^2 √(a^2-x^2) dx from 0 to a


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The Attempt at a Solution



So i don't need actual help with this problem. I got the answer, (π*a^4)/16 and I verified with the back of the book. The question I have is whether this problem merits an entire side of work? None of the examples my professor has given have ever been more than a few lines of work and this took me a whole side of a paper. Am I being inefficient or should I just expect this from now on?

Oh and sorry if my notation bad or if this should be on another thread, this is my first post lol .__.
 
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I guess it amounts to how much you write. It requires a trig substitution followed by a double angle formula. Is that how you did it? Did you carry the limits along with the substitution or did you back substitute to x? That takes more steps. I used about 1/2 of one side of a standard sheet of paper for it.
 
I decided to back substitute into x, i thought converting the limits would be a hassle on this question. I think my issue was getting a sin4θ and not knowing any quick identities to simplify. Oh well, thanks for the response
 
You could have changed the integration limits after substitution. If you substituted x/a=sin(u) then the integral with respect to u goes from 0 to pi/2.

ehild
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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