Trig substitutions

Our motivation for this section is the following integral

\int \frac{\sqrt{25x^2 - 4}}{x} \, dx

This integral is very difficult to solve with the previous techniques. The trick comes from the substitution

x = \frac{2}{5}\sec \theta

Which after a fair amount of algebra and calculus gives us

\int \frac{\sqrt{25x^2 - 4}}{x} \, dx = \sqrt{25x^2 - 4} + 2\arccos \left(\frac{2}{5x}\right) + C

Solving these integrals relies on a method known as trigonometric substitutions.

Method

First, lets explain how we solved the previous integral.

Example

We have the following integral, which is difficult to solve using previous techniques.

\int \frac{\sqrt{25x^2 - 4}}{x} \, dx

Using the substitution

x = \frac{2}{5}\sec \theta

We can reduce the integral and get a solution in terms of \theta.

\begin{array}{ll} & = \displaystyle \int \frac{\sqrt{25(\frac{2}{5}\sec \theta)^2 - 4}}{\frac{2}{5}\sec \theta} \, (\frac{2}{5}\tan \theta \sec \theta \, d\theta)\\ & = \displaystyle\int \sqrt{4\sec^2 \theta - 4} \, \tan \theta \, d\theta\\ & = \displaystyle2\int \sqrt{\sec^2 \theta - 1} \, \tan \theta \, d\theta = 2\int\tan^2\theta \, d\theta\\ & = 2\left(\tan \theta + \theta\right) + C \end{array}

Now we only need to undo our substitution. Notice that

\sec \theta = \frac{\text{hyp}}{\text{adj}} = \frac{5x}{2}

This can be visualized in the triangle.

Code 
\usetikzlibrary{graphs,graphs.standard,angles,quotes}

\begin{tikzpicture}

\draw[fill=black] (0,0);
\draw[fill=black] (3,0);
\draw[fill=black] (3,2);
\draw[line width=1pt] (0,0) -- (3,0) -- (3,2) -- (0,0);
\node at (1.75,-.25) {$2$};
\node at (1.2,1.25) {$5x$};
\node at (3.9,1) {$\sqrt{25x^2 - 4}$};

\coordinate (origo) at (0,0);
\coordinate (pivot) at (3,2);
\coordinate (mary) at (3,0);

\pic[draw, -, "$\theta$", angle eccentricity=1.5]{angle = mary--origo--pivot};
\end{tikzpicture}

From here can see that \tan \theta = \frac{\sqrt{25x^2-4}}{2}

\theta = \arccos \left(\frac{2}{5x}\right)

Giving us our final answer

\int \frac{\sqrt{25x^2 - 4}}{x} \, dx = \sqrt{25x^2-4} + 2\arccos \left(\frac{2}{5x}\right) + C

When faced with an integral containing a radical expression in the form \sqrt{\pm(bx + c)^2 \pm a}, we can usually find some subtitution for x which simplifies the expression into a singular trigonometric identity. This is achievable due to the following three identities.

\begin{array}{ll} 1 - \sin^2x &=& \cos^2 x\\ \tan^2x + 1 &=& \sec^2 x\\ \sec^2x - 1 &=& \tan^2 x \end{array}

Lets see how these identites allow us to reduce certain radicals.

Lets examine the three cases where trig identities allow us to simplify a radical expression.

Case I: \sqrt{a - (bx + c)^2}

Notice how this case is a constant, subtracted by a squared variable term. This resembles our first expression, 1 - \sin^2x = \cos^2 x. We want to get the expression in these terms so we can reduce the radicle. Notice by making the substitution

(bx + c) = \sqrt{a}\sin \theta

We get the equation

\sqrt{a - (\sqrt{a}\sin\theta)^2} = \sqrt{a}\sqrt{1-\sin^2\theta} = \sqrt{a}\cos\theta

We have now elimated the variable from the radical.

Case II: \sqrt{(bx + c)^2 + a}

This case resembles our identity \tan^2x + 1 = \sec^2 x. So lets try the substitution

(bx + c) = \sqrt{a}\tan \theta

Which gives us

\sqrt{(\sqrt{a}\tan \theta)^2 + a} = \sqrt{a}\sqrt{\tan^2 \theta + 1} = \sqrt{a}\sec\theta

We have now elimated the variable from the radical.

Case III: \sqrt{(bx + c)^2 - a}

This case resembles \sec^2x - 1. Using the substitution

(bx + c) = \sqrt{a}\sec \theta

We get the equation

\sqrt{(\sqrt{a}\sec \theta)^2 - a} = \sqrt{a}\sqrt{\sec^2\theta - 1} = \sqrt{a}\tan\theta

We have now elimated the variable from the radical.

For ease of use, the following table shows what substitution to make depending on the case.

\begin{array}{ll} \text{Radical expression} \hspace{0.25in} & \text{Substitution}\\ &\\ \sqrt{a - (bx + c)^2} \hspace{0.25in} & (bx + c) = \sqrt{a}\sin\theta\\ &\\ \sqrt{(bx + c)^2 + a} \hspace{0.25in} & (bx + c) = \sqrt{a}\tan\theta\\ &\\ \sqrt{(bx + c)^2 - a} \hspace{0.25in} & (bx + c) = \sqrt{a}\sec\theta \end{array}

Examples

Problem

Evaluate the following integral

\int \frac{\sqrt{9 - x^2}}{x^2}\,dx

Notice that the radicle expression resembles the third row of our table. If we make the substitution

x = 3\sin \theta\\ dx = 3\cos\theta\,d\theta

We get

\int \frac{\sqrt{9 - x^2}}{x^2}\,dx = \int\frac{3\sqrt{1-\sin^2\theta}}{9\sin^2\theta}(3\cos\theta)\,d\theta = \int\frac{\cos^2\theta}{\sin^2\theta} d\theta

Which evaulates to \int\frac{\cos^2\theta}{\sin^2\theta} d\theta = -\cot\theta - \theta + C

This however, is only half the problem. We still need to convert the equation back to its orginal variable. Once again, the method of drawing a triangle will help to simplify the final expression.

Note that \sin \theta = \displaystyle\frac{\text{opp}}{\text{hyp}} = \displaystyle\frac{x}{3}

Code 
\usetikzlibrary{graphs,graphs.standard,angles,quotes}

\begin{tikzpicture}

\draw[fill=black] (0,0);
\draw[fill=black] (3,0);
\draw[fill=black] (3,2);
\draw[line width=1pt] (0,0) -- (3,0) -- (3,2) -- (0,0);
\node at (1.75,-.3) {$\sqrt{9 -x^2}$};
\node at (1.2,1.25) {$3$};
\node at (3.2,1) {$x$};

\coordinate (origo) at (0,0);
\coordinate (pivot) at (3,2);
\coordinate (mary) at (3,0);

\pic[draw, -, "$\theta$", angle eccentricity=1.5]{angle = mary--origo--pivot};
\end{tikzpicture}

So

\cot \theta = \frac{\text{adj}}{\text{opp}} = \frac{\sqrt{9 -x^2}}{x}

Giving us a final answer of

\int \frac{\sqrt{9 - x^2}}{x^2}\,dx = \frac{-\sqrt{9-x^2}}{x} - \arcsin\left(\frac{x}{3}\right) + C