calculuspowerup.com

Integration by Trigonometric Substitution

April 6th, 2009 |

Author: Merrin

The integration of a set of three irrational forms can often be handled by a trigonometric substitution. These forms are the following

$(b^{2} -x^{2} )^{n+1/2} \qquad (b^{2} +x^{2} )^{n+1/2} \qquad (x^{2} -b^{2} )^{n+1/2}$

The goal behind the substitution is to eliminate the radical. For example when we plug

$x=b\sin \theta$

into the first form we get

$(b^{2} -b^2\sin^2 \theta )^{n+1/2} =b^{2n+1} (\cos \theta )^{2n+1}$

When we plug

$x=b\tan \theta$

into the second form we get

$(b^2 +b^2 \tan^2 \theta )^{n+1/2} =b^{2n+1} (\sec \theta )^{2n+1}$

When we plug

$x=b\sec \theta$

into the third form we get

$(b^{2} \sec^2 \theta -b^2 )^{n+1/2} =b^{2n+1} (\tan \theta )^{2n+1}$

Furthermore if the relation for dtheta in terms of dx doesn’t pose any problems we know how to integrate all the powers of different combinations of sine and cosine. Those integrals we did will be pretty useful after all for integrals containing these types of irrational forms.

Here are some nontrivial examples involving the irrational forms. By nontrivial, we don’t know these integrals by our table of antiderivatives.

$\displaystyle \int \cfrac{dx}{x^{2} \sqrt{x^{2} -b^{2} } } \quad \quad \quad \quad \, \displaystyle \int \sqrt{b^{2} -x^{2} } \quad \quad \quad \quad \displaystyle \int \cfrac{dx}{\sqrt{b^{2} +x^{2} } }$

Before we try these we should take a look at the integrals we already know and solve them a second way.

$\int \cfrac{dx}{\sqrt{1-x^{2} } } =\text{arcsin}\, x+C } \\ {\displaystyle \int \cfrac{dx}{1+x^{2} } =\text{arctan}\, x+C} \\ {\displaystyle \int \cfrac{dx}{x\sqrt{x^{2} -1} } =\text{arcsec}\, x+C$

We can solve these last three integrals without reference to any tables, by using trigonometric substitutions. The substitution for the integral that is

$\text{arcsin}\,x$

is

$x=\sin \theta$

and so on for the other integrals.

Example 1. Solve the three fundamental trigonometric substitution integrals.

Solution 1. For the first integral we have

$\displaystyle \int \cfrac{dx}{\sqrt{1-x^{2} } }$

The recommended substitution for the above integral is

$x=\text{sin}\,\, \theta \quad \, dx=\text{cos}\,\, \theta d\theta$
$\displaystyle \int \cfrac{dx}{\sqrt{1-x^{2} } } =\displaystyle \int \cfrac{\text{cos}\,\, \theta d\theta }{\sqrt{1-\text{sin}\,\, ^{2} \theta } } =\displaystyle \int d\theta =\theta +C=\text{arcsin}\,\, x+C$

For the second integral, we have

$\displaystyle \int \cfrac{dx}{1+x^{2} }$

The recommended substitution above is

$x=\text{tan}\,\, \theta \quad \, dx=\text{sec}\,\, ^{2} \theta d\theta$
$\displaystyle \int \cfrac{dx}{1+x^{2} } =\displaystyle \int \cfrac{\text{sec}\,\, ^{2} \theta d\theta }{1+\text{tan}\,\, ^{2} \theta } =\displaystyle \int d\theta =\theta +C=\text{arctan}\,\, x+C$

For the third integral we have

$\displaystyle \int \cfrac{dx}{x\sqrt{x^{2} -1} }$

The recommended substitution above is

$x=\text{sec}\,\, \theta \quad \, dx=\text{sec}\,\, \theta \text{tan}\,\, \theta d\theta$

$\displaystyle \int \cfrac{dx}{x\sqrt{x^{2} -1} } =\displaystyle \int \cfrac{\text{sec}\,\, \theta \text{tan}\,\, \theta d\theta }{\text{sec}\,\, \theta \sqrt{\text{sec}\,\, ^{2} \theta -1} } =\displaystyle \int d\theta =\theta +C=\text{arcsec}\,\, x+C$

Now that we can solve the basics let us try to solve some real problems. But before we do, we need to know how to do some intermediate calculations. When you solve trigonometric integral problems sometimes the integral ends up having trigonometric functions in the result. You have to be able to convert these results in your substituted variables back into a function of \textit{x}. You can make this transformation with the aid of what is called a reference triangle.

You draw a right triangle where two of the side lengths are known from the nature of the substitution. The third side length can be found from the Pythagorean Theorem. Then any trigonometric quantity can be calculated in terms of these side lengths which are functions of \textit{x}.

\begin{figure}[h]
\begin{center}
\includegraphics[width= 3in]{../images/trigref2.pdf}
\caption{A trigonometric reference triangle}
\end{center}
\end{figure}
Once you have drawn the reference triangle it is easy to express the angle theta in terms of the sides of the triangle many different ways.

${\theta }={\text{arcsin}\,\, (y/\sqrt{x^{2} +y^{2} } )=\text{arccos}\,\, (x/\sqrt{x^{2} +y^{2} } )=\text{arctan}\,\, (y/x)}$

$={\text{arccsc}\,\,(\sqrt{x^{2} +y^{2} } /y)= \text{arcsec}\,\,(\sqrt{x^{2} +y^{2} } /x)=\text{arccot}\,\,(x/y)}$

There are three suggested trigonometric substitutions which give you different reference triangles.

trigref3

Three trigonometric reference triangles correspond to the recommended substitutions.

It is easy to use these triangles. For the substitution you are doing such as x = a sin t and so on you can read off trigonometric functions of the angle t by using opposite, adjacent, and hypotenuse. For example when you do a substitution x = a tan t then cos t = adjacent over hypotenuse for the middle triangle a/sqrt(a^2+x^2).

Now we will move on to calculate our three examples.

$\displaystyle \int \cfrac{dx}{x^{2} \sqrt{x^{2} -a^{2} } } \quad \quad \quad \quad \, \displaystyle \int \sqrt{a^{2} -x^{2} } \quad \quad \quad \quad \displaystyle \int \cfrac{dx}{\sqrt{a^{2} +x^{2} } }$

The middle integral looks the simplest because everything is in the numerator so we will try that one first.

Example 2. Solve the integral

$\displaystyle \int \sqrt{b^{2} -x^{2} } dx$

Solution 2. We use the recommended substitution for this form,

$x=a\sin \theta \quad dx=a\cos \theta d\theta$

$\int \sqrt{a^{2} -x^{2} } dx } {=} {\displaystyle \int \sqrt{a^{2} -a^{2} \text{sin}\,\, ^{2} \theta } a\text{cos}\,\, \theta d\theta = a^{2} \displaystyle \int \text{cos}\,\, ^{2} \theta d\theta$

$(a^2/2)\displaystyle \int(1+\text{cos}\,\,2\theta)d\theta= (a^2/2)(\theta + \text{sin}\,\,(2\theta)/2)+C$

$I=(a^2/2)(\theta+\text{sin}\,\,(\theta)\text{cos}\,\,(\theta))+C$

Completing the integration requires the use of our reference triangle three times.

$I=\cfrac{1 }{2} \left(a^2\text{arcsin}\,\, (x/a)+x\sqrt{a^{2} -x^{2} } \right)$

Example 3. Solve another of the nontrivial examples. Find the integral of

$\quad \displaystyle \int \cfrac{dx}{\sqrt{a^{2} +x^{2} } }$

Solution 3. The recommended substitution to use is

$x=a\text{tan}\,\, \theta \quad \, dx=a\text{sec}\,\, ^{2} \theta d\theta$
$\displaystyle \int \cfrac{dx}{\sqrt{a^{2} +x^{2} } } \to \displaystyle \int \cfrac{\text{sec}\,\, ^{2} \theta d\theta }{a\sqrt{1+\text{tan}\,\, ^{2} \theta } } =\cfrac{1}{a} \displaystyle \int \text{sec}\,\, \theta d\theta$

$I=\cfrac{1}{a} \ln \left|\text{sec}\,\, \theta +\text{tan}\,\, \theta \right| +C$

To complete the integral we have to use our reference triangle to evaluate secant and tangent.

$\text{tan}\,\, \theta = \cfrac{x}{a} \qquad \text{sec}\,\, \theta = \sqrt{x^2+a^2}{a}$

$\displaystyle \int \cfrac{dx}{\sqrt{a^{2} +x^{2} } } =\ln \left|\cfrac{\sqrt{x^{2} +a^{2} } }{a} +\cfrac{x}{a} \right|+C_1= \ln \left| \sqrt{x^2 +a^2} + x \right|+C$

Example 4. Solve the following integral by trigonometric substitution.

$\displaystyle \int \cfrac{dx}{x^{2} \sqrt{x^{2} -b^{2} } }$

Solution 4. The recommended substitution for this form is

$x=a\text{sec}\,\, \theta \quad dx=b\text{sec}\,\, \theta \text{tan}\,\, \theta d\theta$
$\displaystyle \int \cfrac{dx}{x^{2} \sqrt{x^{2} -b^{2} } } \to \displaystyle \int \cfrac{b\sec \theta \tan \theta d\theta }{b^{2} \sec ^{2} \theta \sqrt{\sec^{2} \theta -1} } =\cfrac{1}{b^{2} } \displaystyle \int \cos \theta d\theta$

$I = \cfrac{1}{b^{2} } \sin \theta +C$

A brief consultation of our reference triangle and we see that

$\displaystyle \int \cfrac{dx}{x^{2} \sqrt{x^{2} -b^{2} } } =\cfrac{\sqrt{x^{2} -b^{2} } }{b^{2} x} +C$

Example 5. What are the most general arcsine and arctangent integrals? These are two commonly occurring integrals

Solution 5. These integrals are quite common.

$\int \cfrac{dx}{\sqrt{a^{2} -(x-b)^{2} } } =\text{arcsin}\,\left(\cfrac{x-b}{a} \right)+C } \\ {\displaystyle \int \cfrac{dx}{a^{2} +(x-b)^{2} }=\cfrac{1}{a} \text{arctan}\, \left(\cfrac{x-b}{a} \right)+C$