Solving the integral $\int_{-1}^1 2\sqrt{2-2x^2}\,dx$ I'm working on a triple integral and have managed to get it to a certain point: 
$$\int_{-1}^1 2\sqrt{2-2x^2}dx $$ 
When I check this with WolframAlpha it gives the answer $\pi\sqrt{2}$, which is the right answer to the problem.
I know I should do a trig substitution to solve from here, so I used $x=\sin t$, which gives me 
$$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{2\cos^2t}\,dt $$
but this has the answer $4\sqrt{2}$. 
I'm probably making a glaringly obvious mistake, but if someone could help me out, I'd greatly appreciate it!
 A: $$\int 2\sqrt{2-2x^2}\, dx=2\sqrt{2}\int \sqrt{1-x^2}\, dx$$
For the expression to make sense, we must have $1-x^2\ge 0$, i.e. $-1\le x\le 1$, so exists $t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ such that $x=\sin t$. Then $dx=\cos t\, dt$.
$$=2\sqrt{2}\int\cos^2 t\, dt=2\sqrt{2}\int \frac{1+\cos(2t)}{2}\, dt$$
$$=\sqrt{2}\left(\int dt+\frac{1}{2}\int \cos(2t)\, d(2t)\right)$$
$$=\sqrt{2}\left(t+\frac{1}{2}\sin(2t)\right)+C=\sqrt{2}\left(\arcsin x+\sin t\cos t\right)+C$$
$$=\sqrt{2}\left(\arcsin x+x\sqrt{1-x^2}\right)+C$$
$$\int_{-1}^12\sqrt{2-2x^2}\, dx=\sqrt{2}\left(\arcsin (1)+1\sqrt{1-1^2}\right)-\sqrt{2}\left(\arcsin (-1)+(-1)\sqrt{1-(-1)^2}\right)$$
$$=\sqrt{2}\left(\arcsin (1)-\arcsin (-1)\right)=\sqrt{2}\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)=\pi\sqrt{2}$$
A: your substitution is wrong, probably you forgot to substitute $dx = \cos{t}\ dt$
$\int_{-\pi/2}^{\pi/2} 2\sqrt{2} \cos{t}.\cos{t}dt = \int_{-\pi/2}^{\pi/2} \sqrt{2} (\cos{2t}+1)dt = \sqrt{2}*(\pi/2+\pi/2)=\pi\sqrt{2}$
A: $f(x)=2\sqrt{2-2x^2}=2\sqrt2 \sqrt{1-x^2}\\\int_{-1}^{1}f(x)dx=\\2\sqrt2\int_{-1}^{1}\sqrt{1-x^2}dx$
or $$2\sqrt2\int_{-1}^{1}\sqrt{1-x^2}dx=\\$$take $x=sint$ so $\\dx=cost dt$ 
$$ 2\sqrt2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{cos^2t}cost dt=\\2\sqrt2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|cost|cost dt=\\2\sqrt2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}cost cost dt=\\2\sqrt2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}cos^2t dt=\\2\sqrt2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\frac{1+cos(2t)}{2}) dt=\\2\sqrt2(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\frac{1}{2}) dt+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\frac{cos(2t)}{2}) dt)=\\2\sqrt2 \times(\frac{1}{2}(\frac{\pi}{2}-(-\frac{\pi}{2}))+0)=\\2\sqrt2 \times(\frac{\pi}{2})$$
A: As pointed out we have
$$
    \int_{-1}^1 2\sqrt{2-2x^2}\, dx=2\sqrt{2}\int_{-1}^1 \sqrt{1-x^2}\, \mathrm{d}x
$$
The crux now is to note that $y = \sqrt{1-x^2} \ \Rightarrow \ y^2 + x^2 = 1^2$.
In other words this integral reprsesents the upper half of a circle with radius 1. The area of a circle with radius $1$ is simply $A = \pi r^2 = \pi$ and so the upper half has an area $\pi/2$. Now the integral is simply
$$
    \int_{-1}^1 2\sqrt{2-2x^2}\, dx=2\sqrt{2}\int_{-1}^1 \sqrt{1-x^2}\, \mathrm{d}x = 2 \sqrt{2} \cdot \frac{\pi \cdot 1^2}{2} = \pi \sqrt{2}
$$
As wanted. The figure shows the the function. 
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