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.

$$\int \sqrt{8 - x^2} dx$$
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.
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), 
$$I=x\sqrt{8-x^2}-I+8\int \frac{1}{\sqrt{(2\sqrt 2)^2-x^2}}\ dx$$
$$2I=x\sqrt{8-x^2}+8\sin^{-1}\left(\frac{x}{2\sqrt 2}\right)+c$$
$$I=\color{red}{\frac{1}{2}\left(x\sqrt{8-x^2}+8\sin^{-1}\left(\frac{x}{2\sqrt 2}\right)\right)+C}$$