TITLE: Is there any way to solve integral of $\sqrt{8-x^{2}}$ without using $\sin$ or $\cos$ formulas?
QUESTION [5 upvotes]: 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.
$$\int \sqrt{8 - x^2} dx$$

REPLY [7 votes]: 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.