# Indefinite integral of $\sqrt{x^2-x}$

i was trying to compute the indefinite integral: $$\int\sqrt{x^2-x}dx$$ but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts: $$\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\ =x\sqrt{x^2-x}-\frac{1}{2}\int x\frac{2x-1}{\sqrt{x^2-x}}dx= \\ =x\sqrt{x^2-x}-\int \frac{x^2}{\sqrt{x^2-x}}dx+\frac{1}{2}\int \frac{x}{\sqrt{x^2-x}}dx=...$$ and now what? Can anybody help?

• Looking at the integral, you might want to make some sort of trigonometric substitution. Commented May 22, 2018 at 0:34
• First, complete the square and let $x-\frac 12=\frac {\sec u}2$ Commented May 22, 2018 at 0:36
• Learn this and then don't ever have any problem with those types of integrals.
– user561348
Commented May 22, 2018 at 0:36
• @FrankW. I found this too from WolframAlpha, yet I am having a hard time finding out how to think of this naturally. Experience, I guess. Commented May 22, 2018 at 0:40
• @ArnaudMortier If you complete the square $\sqrt{x^2-x}=\frac{(2x-1)^2-1}{2}$. Then $\sin^2+\cos^2=1$ gives $\frac{1}{\cos^2}-1=\frac{\sin^2}{\cos^2}$. This and $\sin^2=1-\cos^2$ are the identities that trigonometric substitutions exploit to simplify the integral (and also their derivatives). Unfortunately, lazy teaching has made trigonometric substitutions more popular than they deserve to be. Or better said, lazy teaching doesn't make Euler's substitutions as popular as they should be, and then the trigonometric one just fill up the void.
– user561348
Commented May 22, 2018 at 0:47

hint

$$\sqrt {x^2-x}=\frac {1}{2}\sqrt {(2x-1)^2-1}$$

then put $$2x-1=\cosh (t)$$ and use

$$\cosh^2(t)-1=\sinh^2(t)$$

• what is $h(t)$ ? Commented May 22, 2018 at 1:03
• @olgchar $\cosh(t)$ and $\sinh(t)$ are the hyperbolic cosine and sine functions. There's more information on the Wikipedia page. Commented May 22, 2018 at 1:20
• aahh ok... silly question! thanks anyway! Commented May 22, 2018 at 1:21

$$\int \sqrt{x^2-x} \ dx$$ $$=\int \sqrt{(x-\frac12)^2-\frac14}dx$$ Apply u-substitution:$$u=x-\frac12$$ $$=\int\sqrt{u^2-\frac14}du$$ Apply Trig Substitution: $$u=\frac12 \sec(t)$$ $$=\int\frac{\sec(t)\tan(t)\sqrt{\sec^2(t)-1}}{4}dt$$ $$\frac14\int \sec(t)\tan(t)\sqrt{\sec^2(t)-1}dt$$ Use the identity:$$\sec^2(x)=1+\tan^2(x)$$ $$\frac14 \int \sqrt{-1+1+\tan^2(t)}(\sec(t)\tan(t))dt$$ $$\frac14 \int \sqrt{\tan^2(t)}(\sec(t)\tan(t))dt$$ $$=\frac14 \int \tan(t)\sec(t)\tan(t)dt$$ $$=\frac14 \int \tan^2(t)\sec(t)dt$$ Using the identity:$$\tan^2(x)=-1+\sec^2(x)$$ $$=\frac14 \int (-1+\sec^2(t))\sec(t)dt$$ $$\frac14 \int -\sec(t)+\sec^3(t)dt$$ $$\frac14(-\int \sec(t)dt+\int \sec^3(t)dt)$$ Note that $$\int \sec(t)dt=\ln |\tan(t)+\sec(t)|$$

and $$\int \sec^3(t)dt=\frac{\sec^2(t)\sin(t)}{2}+\frac12 \ln |\tan(t)+\sec(t)|$$ $$=\frac14(-\ln |\tan(t)+\sec(t)|+\frac{\sec^2(t)\sin(t)}{2}+\frac12 \ln |\tan(t)+\sec(t)|)$$ And finally after substituting back the "u and t", we get $$\int \sqrt{x^2-x}dx=\frac18 (4x^2 \sqrt{-\frac{1}{(2x-1)^2}+1}-4x \sqrt{-\frac{1}{(2x-1)^2}+1}+ \sqrt{-\frac{1}{(2x-1)^2}+1}-\ln|2x\sqrt{-\frac{1}{(2x-1)^2}+1}-\sqrt{-\frac{1}{(2x-1)^2}+1}+2x-1|)+C$$

complete the square

\begin{align} \int\sqrt{x^2-x} \ dx &=I \\ x^2-x&=\left(x-\frac{1}{2}\right)^2-\frac{1}{4} \\ \text{set} \quad u&=x-\frac{1}{2} \quad \text{then} \quad dx=du\\ I&=\int\sqrt{u^2-a} \ du \quad \text{such that} \quad a=\frac{1}{4}\\ \end{align}

This type of integral is well known. You should now use a trig substitution.

$$\int\sqrt{x^2-x}\; dx=\int\sqrt{x}\; \sqrt{x-1}\; dx\qquad\qquad x\rightarrow\; \cosh^2\theta\quad dx\rightarrow2\cosh\theta\sinh\theta\; d\theta$$

$$=2\int\cosh^2\theta\sinh^2\theta\; d\theta\ =2\int\cosh^2\theta(\cosh^2\theta-1)\; d\theta\ = 2\int\cosh^4\theta\; d\theta -2\int\cosh^2\theta\; d\theta\$$

$$=\frac{1}{2}\int(1+\cosh(2\theta))^2\;d\theta\ -\int(1+cosh(2\theta))\; d\theta$$ $$\frac{1}{2}\int(1+2\cosh(2\theta)+\cosh^2(2\theta))\; d\theta\ -\theta-\frac{1}{2}\sinh(2\theta)+C$$ $$\frac{\theta}{2}+\frac{1}{2}\sinh(2\theta)-\theta-\frac{1}{2}\sinh(2\theta)+C\ +\frac{1}{4}\int(1+\cosh(4\theta))\; d\theta$$ $$\frac{\theta}{2}+\frac{1}{2}\sinh(2\theta)-\theta-\frac{1}{2}\sinh(2\theta)\ +\frac{\theta}{4}+\frac{1}{16}\sinh(4\theta)+C$$ $$\frac{1}{16}\sinh(4\theta)-\frac{\theta}{4}+C$$ $$\theta=\cosh^{-1}(\sqrt{x})$$ $$\frac{1}{16}\sinh(4\cosh^{-1}(\sqrt{x}))-\frac{1}{4}\cosh^{-1}(\sqrt{x})+C$$ This is the short answer if you don't want so much details. Of course if you have enough skills can try using Integration By Parts or another Trig. Subs.

Integrate by parts instead as follows \begin{align}\int\sqrt{x^2-x}\ dx = &\int\frac{\sqrt{x^2-x}}{2(x-\frac12)}d[(x-\frac12)^2]\\ = &\ \frac12(x-\frac12) \sqrt{x^2-x}-\frac18\int \frac 1{ \sqrt{x^2-x}}dx \\ =& \ \frac12(x-\frac12) \sqrt{x^2-x} -\frac18\tanh^{-1}\frac{\sqrt{x^2-x}}{x-\frac12} \end{align}

We can eliminate the square root with an Euler substitution of

$$t = \sqrt{x^2-x} - x \implies x = -\frac{t^2}{1+2t} \implies dx = -\frac{2t(1+t)}{(1+2t)^2} \, dt$$

Then

\begin{align*} I &= \int \sqrt{x^2-x} \, dx \\[1ex] &= -2 \int \left(t-\frac{t^2}{1+2t}\right) \frac{t(1+t)}{(1+2t)^2} \, dt \\[1ex] &= -2 \int \frac{t^2(1+t)^2}{(1+2t)^3} \, dt \\[1ex] &= - \frac18 \int \left(1 + 2t - \frac2{1+2t} + \frac1{(1+2t)^3}\right) \, dt \end{align*}