Calculus – Integral of sqrt(1-x^2) Using Integration by Parts

Question

I was asked to solve this indefinite integral using Integration by parts.

$$\int \sqrt{1-x^2} dx$$

I know how to solve if use the substitution $x=\sin(t)$ but I'm looking for the Integration by parts way.

any help would be very appreciated.

Answer 1

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 $\sin 2t = 2\sin t\cos t = 2x\sqrt{1-x^2}$.