How to choose a trigonometric substitution

A square root with a sum or difference of squares often gets easier when we use a trig identity. Which substitution we choose depends on the expression inside the radical. We’ll look at the three patterns and use one after completing the square.

Match the radical to a trig identity

Choose the substitution that simplifies the square root
Radical	Substitution
$\sqrt{a^{2} - u^{2}}$	$u = a \sin θ$
$\sqrt{a^{2} + u^{2}}$	$u = a \tan θ$
$\sqrt{u^{2} - a^{2}}$	$u = a \sec θ$

Here $a > 0$ . Choose an angle range that keeps the radical nonnegative and makes the substitution reversible on the interval you are using.

Complete the square to find a secant pattern

\int \frac{x}{\sqrt{2 x^{2} - 4 x - 7}} d x

The expression under the root is $2 (x - 1)^{2} - 9$ . Use $u = \sqrt{2} (x - 1)$ , and on the right-hand interval use $u = 3 \sec θ$ with $0 < θ < \frac{π}{2}$ . This gives $\sqrt{u^{2} - 9} = 3 \tan θ$ .

\begin{array}{l} x = 1 + \frac{3}{\sqrt{2}} \sec θ \\ d x = \frac{3}{\sqrt{2}} \sec θ \tan θ d θ \end{array}

After canceling the tangent factor, the integral becomes

\begin{array}{l} \frac{1}{\sqrt{2}} \int \sec θ d θ \\ + \frac{3}{2} \int \sec^{2} θ d θ \end{array}

Return to the original variable

Write $D = 2 x^{2} - 4 x - 7$ . Then $\tan θ = \frac{\sqrt{D}}{3}$ on this branch. Using the two antiderivatives gives

\begin{array}{l} \frac{\sqrt{D}}{2} \\ + \frac{1}{\sqrt{2}} \ln | \sqrt{2} (x - 1) + \sqrt{D} | + C \end{array}

The factor of $3$ inside the logarithm can be absorbed into $C$ . Differentiation confirms this expression on either interval where $D > 0$ . Each interval can have its own constant.

Learn mathKrista KingApril 1, 2019math, learn online, online course, online math, equilibrium solutions

How to choose a trigonometric substitution

Match the radical to a trig identity

Complete the square to find a secant pattern

Return to the original variable

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