Examples

Trig sub in quadratic - completing the square

Compute the integral:

$$\int \frac{dx}{\sqrt{x^2-6x+11}}$$

Solution

(1) Complete the square to obtain Pythagorean form:

$$x^2-6x+{\left(\frac{-6}{2}\right)}^2\gg \gg x^2-6x+9={(x-3)}^2$$

Add and subtract to get desired constant term:

$$\begin{array}{c}{x}^2-6x+11\gg \gg x^2-6x+9-9+11\\ \\ \gg \gg {(x-3)}^2+2\end{array}$$

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(2) Perform shift substitution:

Set $u=x-3$ as inside the square:

$${(x-3)}^2+2=u^2+2$$

Infer $du=dx$. Plug into integrand:

$$\int \frac{dx}{\sqrt{x^2-6x+11}}\gg \gg \int \frac{du}{\sqrt{u^2+2}}$$

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(3) Trig sub with $\tan \theta$:

Use substitution $u=\sqrt{2}\tan \theta$. (From triangle or memorized tip.)

Infer $du=\sqrt{2}{\sec }^2\theta d\theta$. Plug in data:

$$\int \frac{du}{\sqrt{u^2+2}}\gg \gg \int \frac{{\sec }^2\theta }{\sec \theta }d\theta =\int \sec \theta d\theta$$

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(4) Integrate using ad hoc memorized formula:

$$\int \sec \theta d\theta \gg \gg \ln |\tan \theta +\sec \theta |+C$$

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(5) Convert trig back to $x$:

First in terms of $u$, referring to the triangle:

$$\tan \theta =\frac{u}{\sqrt{2}},\sec \theta =\frac{\sqrt{u^2+2}}{\sqrt{2}}$$

Then in terms of $x$ using $u=x-3$. Plug everything in:

$$\ln |\tan \theta +\sec \theta |+C\gg \gg \ln |\frac{x-3}{\sqrt{2}}+\frac{\sqrt{{(x-3)}^2+2}}{\sqrt{2}}|+C$$

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(6) Simplify using log rules:

$$\ln \frac{f(x)}{a}\gg \gg \ln f(x)-\ln a$$

The common denominator $\frac{1}{\sqrt{2}}$ can be pulled outside as $-\ln \sqrt{2}$.

The new term $-\ln \sqrt{2}$ can be “absorbed into the constant” (redefine $C$).

So we write our final answer thus:

$$\ln |x-3+\sqrt{{(x-3)}^2+2}|+C$$