Is there any way to solve integral of $\sqrt{8-x^{2}}$ without using $\sin$ or $\cos$ formulas? I was thinking about the following integral if I could solve it without using trigonometric formulas. If there is no other way to solve it, could you please explain me why do we replace $x$ with $2\sqrt 2 \sin(t)$? I'm really confused about these types of integrals.
$$\int \sqrt{8 - x^2} dx$$
 A: By rescaling the variable, let us replace the constant $8$ by $1$, for convenience.
The equation $y=\sqrt{1-x^2}$ represents the upper-half of the unit circle, and the integral
$$\int_{t=0}^x\sqrt{1-t^2}dt$$ is the area of a vertical "slice" between the abscissas $0$ and $x$. You can compute it as the area of a sector of aperture $\theta$ such that $\sin(\theta)=x$, plus a triangle of base $x$ and height $\sqrt{1-x^2}$.

Hence,
$$A=\frac12\theta+\frac12x\sqrt{1-x^2}=\frac12\arcsin(x)+\frac12x\sqrt{1-x^2}.$$
This is how a trigonometric function appears, and you can't avoid it because it belongs to the final solution.

You also see the connection by taking the derivative
$$(\arcsin(x))'=\frac1{\sqrt{1-x^2}}.$$
The trigonometric function disappears and is replaced by a rational expression.
A similar phenomenon occurs with the logarithm,
$$(\ln(x))'=\frac1x,$$
and this is why you will see logarithms appear now and then in antiderivatives.
A: The idea is that we want to get rid of the square root. So we use the fact that $1 - \sin^2 = \cos^2$, so that $$\sqrt{1 - \sin ^2 x } = \sqrt{\cos^2 x} = |\cos x|$$
which we then can integrate (of course there will be another factor coming from $dx$ but that works out)
Same idea when you're faced with $\sqrt{1 + x^2}$; this time we use the fact that $1 + \sinh^2 = \cosh^2$ and we get rid of the square root in the same way.
Of course if you have a number $a$ instead of $1$, you need to be able to factor that; so you want to transform $\sqrt{a - x^2}$ in $\sqrt{a - a\sin^2 x} = \sqrt a \sqrt{1 - \sin^2 x} =\sqrt  a |\cos x|$
A: Let
$$I=\int\sqrt{8-x^2}\ dx\tag 1$$
using integration by parts,
$$I=\sqrt{8-x^2}\int 1\ dx-\int \left(\frac{-2x}{2\sqrt{8-x^2}}\right)\cdot x\ dx$$
$$I=\sqrt{8-x^2}(x)-\int \frac{(8-x^2)-8}{\sqrt{8-x^2}} \ dx$$
$$I=x\sqrt{8-x^2}-\int \left(\sqrt{8-x^2}-\frac{8}{\sqrt{8-x^2}} \right)\ dx$$
$$I=x\sqrt{8-x^2}-\int\sqrt{8-x^2}\ dx+8\int \frac{1}{\sqrt{8-x^2}}\ dx$$
setting the value from (1), 
$$I=x\sqrt{8-x^2}-I+8\int \frac{1}{\sqrt{(2\sqrt 2)^2-x^2}}\ dx$$
$$2I=x\sqrt{8-x^2}+8\sin^{-1}\left(\frac{x}{2\sqrt 2}\right)+c$$
$$I=\color{red}{\frac{1}{2}\left(x\sqrt{8-x^2}+8\sin^{-1}\left(\frac{x}{2\sqrt 2}\right)\right)+C}$$
