Integral of $\sqrt{1-x^2}$ using integration by parts

I'll restate the accepted answer in different notation, which is easier for me to parse: let $$u=\sqrt{1-x^2}, \quad dv=dx$$ so that $$du=\frac{-x}{\sqrt{1-x^2}}dx,\quad v=x$$ For brevity, write $I=\int \sqrt{1-x^2}\, dx$. Using $\int u\,dv = uv-\int v\,du$, obtain $$ I = x\sqrt{1-x^2} - \int \frac{-x^2}{\sqrt{1-x^2}} \,dx $$ The last integral does not look simpler than $I$ itself, but it can be related back to it: $$ \int \frac{-x^2}{\sqrt{1-x^2}} \,dx = \int \frac{1-x^2}{\sqrt{1-x^2}} \,dx - \int \frac{1 }{\sqrt{1-x^2}} \,dx = I - \sin^{-1}x $$ So, $$I = x\sqrt{1-x^2} - (I-\sin^{-1}x)$$ and solving for $I$ yields $$\int \sqrt{1-x^2}\, dx = \frac12 x\sqrt{1-x^2} + \frac12 \sin^{-1}x + C$$

For completeness and comparison, I'll add the conventional solution using $x=\sin t$ substitution. Here $dx=\cos t\,dt$, so $$ \int\sqrt{1-x^2}\,dx = \int \cos^2 t\,dt =\int \left(\frac12+\frac{\cos 2t}{2}\right)\,dt = \frac{t}{2}+\frac{\sin 2t}{4}+C $$ To return to $x$, note that $t=\sin^{-1}x$ and $\sin 2t = 2\sin t\cos t = 2x\sqrt{1-x^2}$.

A different approach, building up from first principles, without using cos or sin to get the identity, $$\arcsin(x) = \int\frac1{\sqrt{1-x^2}}dx$$ where the integrals is from 0 to z. With the integration by parts given in previous answers, this gives the result.

The distance around a unit circle traveled from the y axis for a distance on the x axis = $\arcsin(x)$.

$$\arcsin(x) = \int\frac{ds}{dx}dx$$ Pythagoras gives the distance in terms of change in y and change in x. $$ds^2 = dx^2 + dy^2$$ $$\frac{ds}{dx} = \sqrt{1 + {\frac{dy}{dx}}^2}$$ x and y are on a unit circle. $$1 = x^2 + y^2$$ Rearranging to solve for y and differentiating. $$\frac{dy}{dx} = \frac{x}{\sqrt{1 - x^2}}$$ Substituting in the above, $$\frac{ds}{dx} = \frac1{\sqrt{1 - x^2}}$$ And subsituting in the integral gives, $$\arcsin(x) = \int\frac1{\sqrt{1-x^2}}dx$$

This integral is much interesting since it can be interpreted in three ways as far as I know.

1. Mentioned by the OP, trigonometric substitution can be applied.
2. Integration by part.
3. Geometric interpretation mentioned in the comment which I think is the most intuitive.

I will give each a reasonable explanation in the following

Trigonometric substitution

Let $x = \sin t$, where $t \in [-\pi / 2, \pi / 2]$ by the domain of the function. \begin{align*} \int \sqrt{1 - x^2} dx &= \int \cos^2 t dt \\ &= \int \frac{1 + \cos 2t}{2} dt\\ &= \frac{1}{2} \left(t + \frac{1}{2} \sin 2t \right) + C \\ &= \frac{1}{2} \left(t + \sin t \cos t \right) + C\\ &= \frac{1}{2} \left(t + \sin t \sqrt{1 - \sin^2 t}\right) + C \end{align*}

Substitute $t$ back to $\sin^{-1} x$, we get \begin{align*} \int \sqrt{1-x^2} dx &= \frac{1}{2} \left(\sin^{-1} x + x \sqrt{1 - x^2}\right) + C. \end{align*}

Integration by part

\begin{align*} \int \sqrt{1 - x^2} dx &= x \sqrt{1 - x^2} - \int x \frac{-x}{\sqrt{1 - x^2}} dx\\ &= x \sqrt{1 - x^2} - \int \frac{1-x^2 - 1}{\sqrt{1 - x^2}} dx\\ &= x \sqrt{1 - x^2} - \int \sqrt{1 - x^2} dx + \int \frac{1}{\sqrt{1 - x^2}} dx \end{align*} Adding \begin{align*} \int \sqrt{1 - x^2} dx \end{align*} to both sides and divide each side by 2, we obtain \begin{align*} \int \sqrt{1 - x^2} dx &= \frac{1}{2} x \sqrt{1 - x^2} + \frac{1}{2} \sin^{-1} x + C. \end{align*}

Geometric interpretation

The original function is a half circle with radius 1, which is seen by \begin{gather*} y = \sqrt{1 - x^2}\\ y^2 = 1 - x^2\\ x^2 + y^2 = 1, \quad y \in [0, 1]. \end{gather*} We are interested in the area formed by $O (0, 0)$, $A (0, 1)$ and $B (x, 0)$ and $C(x, \sqrt{1 - x^2})$ which is the intersection of the vertical line through point $B$ and the half circle. We want to calculate the area of the region $OACB$. The area of the sector formed by $O, A$ and $C$ is easy to calculate \begin{align*} \text{Area of sector $OAC$} &= \frac{1}{2} \left(\frac{\pi}{2} - \cos^{-1} x\right)\\ &= \frac{1}{2} \sin^{-1} x. \end{align*} Area of the triangle $OBC$ is also easy to calculate \begin{align*} \text{Area of $\triangle OBC$} = \frac{1}{2} x \sqrt{1- x^2}. \end{align*} Thus, we obtain the area of the region $OACB$ as \begin{align*} \text{Area of $OACB$} &= \text{Area of sector $OAC$} + \text{Area of $\triangle OBC$}\\ &= \frac{1}{2} \sin^{-1} x + \frac{1}{2} x \sqrt{1 - x^2}, \end{align*} which is one of the antiderivatives of $\sqrt{1 - x^2}$ with the constant $C = 0$.

I tried the geometric approach mentioned in the comments.

Consider the right top quarter of the unit circle with vertices $O(0,0),$$A(1,0),$ and $B(0,1)$. Let $P(x,0)$ and $Q(x,\sqrt{1-x^2})$ on the unit circle. Then, by area calculation, $$ \int_0^x\sqrt{1-x^2}dx=(OPQB)\\ =(OAB)-(OAQ)+(OPQ)\\ =\frac\pi 4-\frac12\angle QOP+\frac12x\sqrt{1-x^2}\\ =\frac\pi 4-\frac12\arccos x+\frac12x\sqrt{1-x^2}\\ =\frac12(\frac\pi 2-\arccos x)+\frac12x\sqrt{1-x^2}\\ =\frac12\arcsin x+\frac12x\sqrt{1-x^2} $$