Tricky Integration with Trig Substitution

bjohnson2001
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Homework Statement



Evaluate.

\int(4-y)\sqrt{4-y^{2}}dy

I have the solution using CAS software here:

2y\sqrt{4-y^{2}}+8sin^{-1}\frac{y}{2}+\frac{1}{3}(4-y^{2})^{3/2}


but I need to do this by hand. I have researched the usual trig methods but am having some difficulty. Can someone please help me find the right identity?
 
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Substituting y=2*sin(t) looks like a good starting point.
 
I don't think it goes anywhere though, I get:

8cos(t)-4sin(t)cos(t)

when it comes time to bring the original variable back in I get a mess of cos(arcsin(y/2)) but maybe I am missing something
 
bjohnson2001 said:
I don't think it goes anywhere though, I get:

8cos(t)-4sin(t)cos(t)

when it comes time to bring the original variable back in I get a mess of cos(arcsin(y/2)) but maybe I am missing something

If y=2*sin(t) then dy=2*cos(t)dt. I think you are forgetting that. And yes, it does take some work to integrate. If you make a try at it and show your steps I'm sure someone will try to help.
 
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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