Trigonometric Substitution in Integrals

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Calculate integrals using trigonometric substitutions with examples and detailed solutions and explanations. Also more exercises with solutions are presented at the bottom of the page.
In all examples and exercises, \( c \) represent the constant of integration.



Trigonometric Identities to Simplify Square Roots

Start with the trigonometric identity
\( \sin^2 x + \cos^2 x = 1 \qquad (I) \)
Rewrite the above identity as
\( 1 - \sin^2 x = \cos^2 x \)
Take the square root of both sides to obtain
\[ \sqrt {1 - \sin^2\;t} = |\cos \; t| \qquad (I') \]

Divide all terms of the identity \( \sin^2 x + \cos^2 x = 1\) by \( \cos x \) to obtain another identity given by
\( \tan^2 x + 1 = \sec^2 x \qquad (II) \)
Take the square root of both sides of the above identity to obtain
\[ \sqrt{\tan^2 x + 1} = |\sec x| \qquad (II') \]

The identity (II) may also be written as
\( \sec^2 x - 1 = \tan^2 x \)
Take the square root of both sides of the above identity to obtain
\[ \sqrt {\sec^2 x - 1} = | \tan x | \qquad (III') \]



A - Calculate Integrals Involving Expressions of the form \( \sqrt {a^2 - b^2 x^2 } \)

Given the expression
\( \sqrt {a^2 - b^2 x^2} \)
Factor \( a^2 \) under the square root and take \( |a| \) out of the square root.
\( \sqrt {a^2 - b^2 x^2} = \sqrt {a^2 \left( 1 - \left(\dfrac{b x}{a}\right)^2\right)} = |a| \sqrt { 1 - \left(\dfrac{b x}{a}\right)^2} \)
Make the substitution
\( \sin t = \dfrac{b x}{a} \)
and rewrite the expression \( \sqrt {a^2 - b^2 x^2 } \) as
\( \sqrt {a^2 - b^2 x^2 } = |a| \sqrt { 1 - \sin^2t} = |a| |\cos t |\)



Example 1
Calculate the integral \[ \displaystyle \int \dfrac{x^2}{\sqrt{16-4x^2}} \; dx \]

Solution to Example 1
Rewrite the expression \( \sqrt{16-4x^2} \) in the denominator of the integrand as
\( \sqrt{16-4x^2} = \sqrt{16(1 - x^2 /4)} = 4 \sqrt{1 - (x/2)^2} \)

Trigonometric substitution: Let \( x/2 = \sin t \) or \( x = 2 \sin t \) which gives \( \dfrac{dx}{dt} = 2 \cos t \) or \( dx = 2 \cos t \; dt \); the integral is given by
\( \displaystyle \int \dfrac{x^2}{\sqrt{16-4x^2}}dx = \int \dfrac{4 \sin^2 t}{4\sqrt{1 - \sin^2 t}} 2 \cos t \; dt \)

Simplify using the identity \( \sqrt { 1 - \sin^2t} = |\cos t | \)
\( = \displaystyle 2 \int \dfrac{ \sin^2 t}{ |\cos t|} \cos t \; dt \)

NOTE that \( |\cos t| \) may only be simplified if we know the sign of \( \cos t \). Since the given integral is indefinite, we may assume that \( \cos t \ge 0 \) and therefore \( |\cos t| = \cos t\). Simplify the above.
\( = \displaystyle 2 \int \sin^2 t \; dt \)

Use the trigonometric identity \( \sin^2 t = (1/2)(1 - \cos 2 t) \); substitute.
\( = \displaystyle \int (1 - \cos 2 t) \; dt \)

Calculate the integral.
\( = \left(t-\dfrac{1}{2}\sin \left(2t\right)\right)+ c \)

Since \( x/2 = \sin t \), \( t = \arcsin (x/2) \). Substitute to obtain the final answer.
\( \displaystyle \int \dfrac{x^2}{\sqrt{16-4x^2}}dx = \arcsin (x/2) -\dfrac{1}{2}\sin \left(2 \arcsin (x/2) \right)+ c \)



B - Calculate Integrals Involving Expressions of the form \( \sqrt {a^2 x^2 + b^2} \)

Given the expression
\( \sqrt {a^2 x^2 + b^2} \)
Factor \( b^2 \) under the square root and take \( |b| \) out of the square root.
\( \sqrt {a^2 x^2 + b^2} = \sqrt {b^2 \left( \left(\dfrac{a x}{b}\right)^2 + 1 \right)} = |b| \sqrt { \left(\dfrac{a x}{b}\right)^2 + 1} \)
Make the substitution
\( \tan t = \dfrac{b x}{a} \)
and rewrite the expression \( \sqrt { a^2 x^2 + b^2 } \) as
\( \sqrt { a^2 x^2 + b^2 } = |b| \sqrt { \tan^2 t + 1} = |b| |\sec t |\)



Example 2
Calculate the integral \[ \displaystyle \int \frac{\sqrt{25x^2+4}}{x^4} \; dx \]

Solution to Example 2
Rewrite the expression \( \sqrt{25x^2+4} \) in the numerator of the integrand as
\( \sqrt{25x^2+4} = \sqrt{4(25 x^2/4 + 1)} = 2 \sqrt{(5 x/2)^2 + 1} \)

Trigonometric substitution: Let \( 5 x/2 = \tan t \) or \( x = (2/5) \tan t \) which gives \( \dfrac{dx}{dt} = (2/5) \sec^2 t \) or \( dx = (2/5) \sec^2 t \; dt \); the integral is given by

\( \displaystyle \int \frac{\sqrt{25x^2+4}}{x^4} \; dx = \int \dfrac{2 \sqrt{\tan^2 t + 1}}{((2/5) \tan t)^4} (2/5) \sec^2 t \; dt \)

Simplify using the identity \( \sqrt { \tan^2 t +1 } = |\sec t | \)
\( \displaystyle = (125/4) \int \dfrac{ |\sec t | }{ \tan^4 t} \sec^2 t \; dt \)

NOTE that \( |\sec t| \) may only be simplified if we know the sign of \( \sec t \). Since the given integral is indefinite, we may assume that \( \sec t \ge 0 \) and therefore \( |\sec t| = \sec t\). Simplify the above integral.
\( \displaystyle = (125/4) \int \dfrac{ \sec^3 t }{ \tan^4 t} \; dt \)

Rewrite the integrand in terms of \( \sin t \) and \( \cos t \)
\( \displaystyle = (125/4) \int \dfrac{ \cos t }{ \sin^4 t} \; dt \)

Substitution: Let \( w = \sin t \) and hence \( \dfrac{d w}{d t} = \cos t \) or \( dt = \dfrac{1}{\cos t} \)

\( \displaystyle = (125/4) \int \dfrac{ 1 }{ w^4 } \; d w \)

Calculate the integral on the right using the power rule.
\( \displaystyle = - (125/4) \dfrac{1}{3 w^3} + c \)

Back substitute knwowing that \( w = \sin t \) and \( t = \arctan( 5 x/2 \) to obtain the final answer.
\( \displaystyle = - (125/12) \dfrac{1}{(\sin(\arctan( 5 x/2))^3} + c \)

Note the we may use the identity \( \sin(\arctan(x)) = \dfrac{x\sqrt{1+x^2}}{1+x^2} \) to simplify the final answer to
\[ \displaystyle \int \dfrac{x^2}{\sqrt{16-4x^2}}dx = - \dfrac{\sqrt{\left(25x^2+4\right)^3}}{ 12 x^3} + c \]



C - Calculate Integrals Involving Expressions of the form \( \sqrt {a^2 x^2 - b^2} \)

Given the expression
\( \sqrt {a^2 x^2 - b^2} \)
Factor \( b^2 \) under the square root and take \( |b| \) out of the square root.
\( \sqrt {a^2 x^2 - b^2} = \sqrt {b^2 \left( \left(\dfrac{a x}{b}\right)^2 - 1 \right)} = |b| \sqrt { \left(\dfrac{a x}{b}\right)^2 - 1} \)
Make the substitution
\( \sec t = \dfrac{b x}{a} \)
and rewrite the expression \( \sqrt { a^2 x^2 - b^2 } \) as
\( \sqrt { a^2 x^2 - b^2 } = |b| \sqrt { \sec^2 t - 1} = |b| |\tan t |\)



Example 3
Calculate the integral \[ \displaystyle \int \dfrac{1}{ \sqrt{4x^2-9}} \; dx \]

Solution to Example 3
Rewrite the expression \( \sqrt{4x^2-9} \) in the denominator of the integrand as
\( \sqrt{4x^2-9} = \sqrt{9(4 x^2/9 - 1)} = 3 \sqrt{(2 x/3)^2 - 1} \)

Trigonometric substitution: Let \( 2 x/3 = \sec t \) or \( x = (3/2) \sec t \) which gives \( \dfrac{dx}{dt} = (3/2) \sec t \tan t \) or \( dx = (3/2) \sec t \tan t \; dt \); the integral is given by

\( \displaystyle \int \dfrac{1}{\left(\sqrt{4x^2-9}\right)} \; dx = \int \dfrac{1}{3 \sqrt{\sec^2 t - 1 }} \; (3/2) \sec t \tan t \; dt \)

Simplify using the identity \( \sqrt { \sec^2 t - 1 } = |\tan t | \)
\( \displaystyle = (1/2) \int \dfrac{1}{|\tan \; t|} \; \sec \; t \tan \; t \; dt \)

NOTE that \( | \tan \;t| \) may only be simplified if we know the sign of \( \tan t \). Since the given integral is indefinite, we may assume that \( \tan t \ge 0 \) and therefore \( |\tan t| = \tan t\). Simplify the above integral.
\( \displaystyle = (1/2) \int \sec t \; dt \)

The integral on the right is a known common integral
\( \displaystyle = (1/2) \ln \left |\tan \; t + \sec \; t \right| + c \)

The substitution \( 2 x/3 = \sec t \) was used and we now need to use \( t = \text{arcsec} (2 x/3) \) in the above to obtain the final answer
\( \displaystyle = (1/2) \ln \; \left|\tan (\text{arcsec} (2 x/3))+2 x/3 \; \right| + c \)

Usng the identity \( \tan \arcsec x =\sqrt{x^2-1} \) to further simplify the final answer

\[ \displaystyle \int \dfrac{1}{\left(\sqrt{4x^2-9}\right)} \; dx = (1/2) \ln \; \left| \sqrt { (2 x/3)^2 - 1 }+2 x/3 \; \right| + c \]



Exercises

Calculate the following integrals.

  1. \( \displaystyle \int \dfrac{x^2}{\sqrt{9-2x^2}} \; dx \)

  2. \( \int \dfrac{\sqrt{9x^2+1}}{x^4} \;dx \)

  3. \( \int \dfrac{1}{\sqrt{9x^2-36}} \; dx \)



Solutions to the Above Exercises

  1. \( \dfrac{9}{4\sqrt{2}} \left(\arcsin \left(\frac{\sqrt{2}}{3}x\right) - \dfrac{1}{2} \sin \left(2\arcsin \left(\frac{\sqrt{2}}{3}x\right)\right)\right)+ c \)

  2. \( -\dfrac{\left(1+9x^2\right)^{\dfrac{3}{2}}}{3x^3}+ c \)

  3. \( \dfrac{1}{3}\ln \left(\dfrac{\left|\sqrt{x^2-4}+x\right|}{2}\right)+C \)

More References and links

  1. integrals and their applications in calculus.

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