Trigonometric Substitutions

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Introduction to this topic[edit]

This page is dedicated to teaching problem solving techniques, specifically for trigonometric substitution. For other integration methods see other sources.

The format is aimed at first introducing the theory, the techniques, the steps and finally a series of examples which will make you further skilled.

Assumed Knowledge[edit]

  • Basic Differentiation
  • Basic Integration Methods
  • Pythagoras Theorem

Theory of Trigonometric Substitutions[edit]

This area is covered by the wikipedia article W:Trigonometric substitution and the wikibooks module B:Calculus/Integration techniques/Trigonometric Substitution. On this page we deal with the practical aspects.
We begin with the following as is described by the above sources.

Trigonometric substitution is a special case of simplifying an intergrand which has a specific form. We will first outline these forms and where they came from.

Pythagoras Theorem[edit]

We should be familiar with pythagoras theorem for a right angled triangle.

a^2 + b^2 = c^2\,

From this familiar definition we can derive other definitions. eg.

 c = \sqrt{a^2 + b^2}. \,

By expanding upon this theory we can come up with other relationships which help us with integration.

Definition 1 Sine Substitution - containing a2x2[edit]

(\sin \theta) = \sqrt{a^2-x^2} \qquad x=a\sin \theta \qquad \sqrt{a^2\cos^2 (\theta)}\,

Trig Sub Triangle 1.png
From the diagram
\sin \theta = \frac {\textrm{opposite}} {\textrm{hypotenuse}} = \frac {x} {a}
(\sin \theta)^2 + x^2 = a^2\,

(\sin \theta)^2 = a^2 - x^2\,
(\sin \theta) = \sqrt{a^2-x^2}\,

x=a\sin \theta\,

\sqrt{a^2-x^2}\,

\sqrt{a^2-(a\sin \theta)^2}\,

\sqrt{a^2-(a^2 \sin^2 (\theta))}\,

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

\cos^2 (\theta) = 1 - \sin^2 (\theta)\,

\sqrt{a^2\cos^2 (\theta)}\,

Definition 2 Tan Substitution - containing a2 + x2[edit]

(\tan \theta) = \sqrt{a^2+x^2} \qquad x=a\tan \theta\,

Trig Sub Triangle 2.png
From the diagram
\tan \theta = \frac {\textrm{opposite}} {\textrm{adjacent}} = \frac {x} {a}
 a^2 + x^2 = (\tan \theta)^2\,

(\tan \theta) = \sqrt{a^2+x^2}\,

x=a\tan \theta\,

\sqrt{a^2+x^2}\,

\sqrt{a^2+(a\tan \theta)^2}\,

\sqrt{a^2+(a^2 \tan^2 (\theta))}\,

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

\sec^2 (\theta) = 1 + \tan^2 (\theta)\,

\sqrt{a^2\sec^2 (\theta)}\,

Definition 3 Sec Substitution - containing x2a2[edit]

 (\sec \theta) = \sqrt{x^2-a^2} \qquad x=a\sec \theta\,

Trig Sub Triangle 3.png
From the diagram
\cos \theta = \frac {\textrm{adjacent}} {\textrm{hypotenuse}} = \frac {a} {x}


\sec \theta = \frac {1} {\cos \theta}

\sec \theta = \frac {\textrm{hypotenuse}} {\textrm{adjacent}} = \frac {x} {a}

 (\sec \theta)^2 + a^2 = x^2 \,

 (\sec \theta)^2 = x^2 - a^2 \,
 (\sec \theta) = \sqrt{x^2-a^2}\,

x=a\sec \theta\,

\sqrt{x^2-a^2}\,

\sqrt{(a\sec \theta)^2-a^2}\,

\sqrt{(a^2\sec^2 \theta)-a^2}\,

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

\sec^2 (\theta) - 1 = \tan^2 (\theta)\,

\sqrt{a^2\tan^2 (\theta)}\,

Summary[edit]

Definition 1 Sine Definition 2 Tan Definition 3 Sec
(\sin \theta) = \sqrt{a^2-x^2}\, (\tan \theta) = \sqrt{a^2+x^2}\, (\sec \theta) = \sqrt{x^2-a^2}\,

This table summarises the definitions that we identify in special integral cases and how they relate to trig identities.

Technique[edit]

Integration 1 Sine Substitution - containing a2x2[edit]

We begin with the integral

\int\frac{dx}{\sqrt{a^2-x^2}}

Step 1 - Identify Trigonometric Substitution Type
We identify this integral as a trigonometric sine substitution.

Step 2 - Identifying Identities for Substitution

x=a\sin(\theta)\,
x\, dx\, \theta\,
x=a\sin(\theta)\, x=a\sin(\theta)\, x=a\sin(\theta)\,
\frac {dx} {d\theta} = a\cos(\theta) \sin(\theta) = \frac {x} {a}
dx=a\cos(\theta)\,d\theta \theta=\arcsin\left(\frac{x}{a}\right)

or
\theta=\sin^{-1} \left(\frac{x}{a}\right)

Step 3 - Substituting Identities into Integral
Now we solve the integral using the following steps

\int\frac{dx}{\sqrt{a^2-x^2}}
= \int\frac{a\cos(\theta)\,}{\sqrt{a^2-a^2\sin^2(\theta)}} \ d\theta \qquad \textrm{substituting} \qquad x=a\sin(\theta)\qquad \textrm{and} \qquad dx=a\cos(\theta)\,d\theta
= \int\frac{a\cos(\theta)\,}{\sqrt{a^2(1-\sin^2(\theta))}} \ d\theta
= \int\frac{a\cos(\theta)\,}{\sqrt{a^2\cos^2(\theta)}} \ d\theta
= \int\frac{a\cos(\theta)\,}{a\cos(\theta)} \ d\theta
= \int d\theta\,
= \theta+C\,

Step 5 - Final Substitution of \theta\,

\text{the question is in terms of } x \text{ so we need the final substitution}\qquad\theta=\arcsin\left(\frac{x}{a}\right)\text{ or } \theta=\sin^{-1} \left(\frac{x}{a}\right)
= \arcsin\left(\frac{x}{a}\right)+C\,\qquad \text{or}
= \sin^{-1}\left(\frac{x}{a}\right)+C\,

Example 1 - Sec substitution[edit]

Evaluate

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


Solution

 \text{In a formal solution there are typically more parts than outlined in the technique section, but the steps remain the same.}\,

Step 1 - Identify Trigonometric Substitution Type

 \text{We look at the format for the square root and recognise it as being a sec substitution.}\,

Step 2 - Identifying Identities for Substitution

 x = 5\sec \theta\,.
x\, x^2-a^2\, dx\, \theta\, \tan\theta\,
x=a\sec(\theta)\, x^2-5^2 = (5\sec\theta)^2-25\, x=5\sec(\theta)\, x=5\sec(\theta)\, x^2-25 = 25\tan^2\theta\,
x=5\sec(\theta)\, = 25(\sec^2\theta - 1)\, \frac {dx} {d\theta} = 5\sec\tan(\theta) \sec(\theta) = \frac {x} {5} \text{solve for }\tan\theta\,
\tan^2\theta\ = \frac {x^2-25}{25}
= 25\tan^2\theta\, dx=5\sec\tan(\theta)\,d\theta \theta=\arcsec\left(\frac{x}{5}\right)

or
\theta=\sec^{-1} \left(\frac{x}{5}\right)

\tan\theta\ = \frac {\sqrt{x^2-25}}{5}\,
 x\,
 x = a\sec \theta \text{.  In this case }a = \sqrt{25} = 5\,\text{ thus }x = 5\sec \theta\,.

x^2-a^2\,
x^2-25 = 25\tan^2\theta\,
x^2-25 = (5\sec\theta)^2-25\,
= 25(\sec^2\theta - 1)\,
= 25\tan^2\theta\,

 dx\,
 dx = 5\sec\tan\theta d\theta\,
 \frac {dx}{d\theta} = 5\sec\tan\theta
\text{NOTE: For differentiating } x= 5\sec \theta \,

x = 5 \sec \theta\,\qquad = 5 \frac {1}{\cos \theta} \qquad = \frac {5}{\cos \theta}\,
 \frac {d} {d\theta} = \frac {vu'-uv'}{v^2}\qquad\text{using Quotient Rule}
 \qquad =\frac {\cos \theta . 0 - 5 . (-\sin \theta)}{\cos^2 \theta}\qquad =\frac {5\sin\theta}{\cos^2\theta} \qquad =\frac {5}{\cos\theta}.\frac{\sin\theta}{\cos\theta}=5\sec\theta\tan\theta

\text{Values for }\theta\,
 \tan\theta\ = \frac {\sqrt{x^2-25}}{5} ,
 \text{recall } x^2-25 = 25\tan^2\theta\,

 \text{solve for }\tan\theta\,

 \tan^2\theta\ = \frac {x^2-25}{25}\,

 \tan\theta\ = \frac {\sqrt{x^2-25}}{5} \,


 \theta = \sec^{-1}\frac{x}{5}\,

 \text{recall } x = 5\sec\theta\,

 \theta = \sec^{-1}\frac{x}{5}\,


Step 3 - Substituting Identities into Integral

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

= \int\frac{\sqrt{25\tan^2\theta}}{5\sec\theta} \ 5\sec\tan\theta \,d\theta\,

= \int\frac{5\tan\theta}{5\sec\theta} \ 5\sec\tan\theta \,d\theta\,

= \int 5\tan\theta \,.\, \tan\theta \,d\theta\,

= 5\int\tan^2\theta \,d\theta\,

= 5\int(\sec^2\theta - 1) \,d\theta\,

= 5(\tan\theta - \theta) + C\,

= 5\tan\theta - 5\theta + C\,

Step 5 - Final Substitution of \theta\,

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

= 5 \,.\, \frac {\sqrt{x^2-25}}{5} - 5 \sec^{-1}\frac{x}{5} + C\,

= \sqrt{x^2-25} - 5 \sec^{-1}\frac{x}{5} + C\,

The Definite Integral[edit]