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, \( c \) represents 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 \( \cos^2 x \) (or rearrange) 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:
\[ \sqrt{a^2 - b^2 x^2} = \sqrt{a^2 \left( 1 - \left(\dfrac{bx}{a}\right)^2 \right)} = |a| \sqrt{1 - \left(\dfrac{bx}{a}\right)^2} \]Make the substitution:
\[ \sin t = \dfrac{bx}{a} \]and rewrite the expression \( \sqrt{a^2 - b^2 x^2} \) as:
\[ \sqrt{a^2 - b^2 x^2} = |a| \sqrt{1 - \sin^2 t} = |a| |\cos t| \]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:
\[ \sqrt{16 - 4x^2} = \sqrt{16\left(1 - \dfrac{x^2}{4}\right)} = 4 \sqrt{1 - \left(\dfrac{x}{2}\right)^2} \]Trigonometric substitution: Let \( \dfrac{x}{2} = \sin t \) or \( x = 2\sin t \), which gives \( \dfrac{dx}{dt} = 2\cos t \) or \( dx = 2\cos t \, dt \); the integral becomes:
\[ \int \dfrac{x^2}{\sqrt{16 - 4x^2}} \, dx = \int \dfrac{(2\sin t)^2}{4\sqrt{1 - \sin^2 t}} (2\cos t) \, dt \]Simplify using the identity \( \sqrt{1 - \sin^2 t} = |\cos t| \):
\[ = 2 \int \dfrac{\sin^2 t}{|\cos t|} \cos t \, dt \]Since the integral is indefinite, we may assume \( \cos t \geq 0 \), so \( |\cos t| = \cos t \):
\[ = 2 \int \sin^2 t \, dt \]Use the trigonometric identity \( \sin^2 t = \dfrac{1}{2}(1 - \cos 2t) \):
\[ = 2 \int \dfrac{1}{2}(1 - \cos 2t) \, dt = \int (1 - \cos 2t) \, dt \]Evaluate the integral:
\[ = t - \dfrac{1}{2}\sin(2t) + C \]Since \( \dfrac{x}{2} = \sin t \), \( t = \arcsin\left(\dfrac{x}{2}\right) \). Substitute back to obtain the final answer:
\[ \int \dfrac{x^2}{\sqrt{16 - 4x^2}} \, dx = \arcsin\left(\dfrac{x}{2}\right) - \dfrac{1}{2}\sin\left(2 \arcsin\left(\dfrac{x}{2}\right)\right) + C \]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:
\[ \sqrt{a^2 x^2 + b^2} = \sqrt{b^2 \left( \left(\dfrac{ax}{b}\right)^2 + 1 \right)} = |b| \sqrt{\left(\dfrac{ax}{b}\right)^2 + 1} \]Make the substitution:
\[ \tan t = \dfrac{bx}{a} \]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 \( \displaystyle \int \dfrac{\sqrt{25x^2 + 4}}{x^4} \, dx \)
Solution to Example 2:
Rewrite the expression \( \sqrt{25x^2 + 4} \) in the numerator as:
\[ \sqrt{25x^2 + 4} = \sqrt{4\left(\dfrac{25x^2}{4} + 1\right)} = 2 \sqrt{\left(\dfrac{5x}{2}\right)^2 + 1} \]Trigonometric substitution: Let \( \dfrac{5x}{2} = \tan t \) or \( x = \dfrac{2}{5}\tan t \), which gives \( dx = \dfrac{2}{5}\sec^2 t \, dt \); the integral is given by:
\[ \int \dfrac{\sqrt{25x^2 + 4}}{x^4} \, dx = \int \dfrac{2\sqrt{\tan^2 t + 1}}{\left(\dfrac{2}{5}\tan t\right)^4} \left(\dfrac{2}{5}\sec^2 t\right) dt \]Simplify using \( \sqrt{\tan^2 t + 1} = |\sec t| \):
\[ = \dfrac{125}{4} \int \dfrac{|\sec t|}{\tan^4 t} \sec^2 t \, dt \]Assuming \( \sec t \geq 0 \), \( |\sec t| = \sec t \):
\[ = \dfrac{125}{4} \int \dfrac{\sec^3 t}{\tan^4 t} \, dt \]Rewrite in terms of \( \sin t \) and \( \cos t \):
\[ = \dfrac{125}{4} \int \dfrac{\cos t}{\sin^4 t} \, dt \]Substitution: Let \( w = \sin t \), so \( dw = \cos t \, dt \):
\[ = \dfrac{125}{4} \int \dfrac{1}{w^4} \, dw \]Calculate using the power rule:
\[ = -\dfrac{125}{12 w^3} + C \]Back substitute \( w = \sin t \) and \( t = \arctan\left(\dfrac{5x}{2}\right) \):
\[ = -\dfrac{125}{12} \cdot \dfrac{1}{\left(\sin\left(\arctan\left(\dfrac{5x}{2}\right)\right)\right)^3} + C \]Using the identity \( \sin(\arctan(x)) = \dfrac{x}{\sqrt{1 + x^2}} \), we simplify to:
\[ = -\dfrac{\sqrt{(25x^2 + 4)^3}}{12x^3} + C \]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:
\[ \sqrt{a^2 x^2 - b^2} = \sqrt{b^2 \left( \left(\dfrac{ax}{b}\right)^2 - 1 \right)} = |b| \sqrt{\left(\dfrac{ax}{b}\right)^2 - 1} \]Make the substitution:
\[ \sec t = \dfrac{bx}{a} \]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 as:
\[ \sqrt{4x^2 - 9} = \sqrt{9\left(\dfrac{4x^2}{9} - 1\right)} = 3 \sqrt{\left(\dfrac{2x}{3}\right)^2 - 1} \]Trigonometric substitution: Let \( \dfrac{2x}{3} = \sec t \) or \( x = \dfrac{3}{2}\sec t \), which gives \( dx = \dfrac{3}{2}\sec t \tan t \, dt \); the integral is:
\[ \int \dfrac{1}{\sqrt{4x^2 - 9}} \, dx = \int \dfrac{1}{3\sqrt{\sec^2 t - 1}} \left(\dfrac{3}{2}\sec t \tan t\right) dt \]Simplify using \( \sqrt{\sec^2 t - 1} = |\tan t| \):
\[ = \dfrac{1}{2} \int \dfrac{1}{|\tan t|} \sec t \tan t \, dt \]Assuming \( \tan t \geq 0 \), \( |\tan t| = \tan t \):
\[ = \dfrac{1}{2} \int \sec t \, dt \]This is a standard known integral:
\[ = \dfrac{1}{2} \ln|\tan t + \sec t| + C \]Since \( \dfrac{2x}{3} = \sec t \), we have \( t = \text{arcsec}\left(\dfrac{2x}{3}\right) \). Substituting back:
\[ = \dfrac{1}{2} \ln\left|\tan\left(\text{arcsec}\left(\dfrac{2x}{3}\right)\right) + \dfrac{2x}{3}\right| + C \]Using the identity \( \tan(\text{arcsec}(x)) = \sqrt{x^2 - 1} \), we simplify to:
\[ = \dfrac{1}{2} \ln\left|\sqrt{\left(\dfrac{2x}{3}\right)^2 - 1} + \dfrac{2x}{3}\right| + C \]Exercises
Calculate the following integrals. Click each exercise to reveal the step-by-step solution.
Exercise 1: Evaluate \( \displaystyle \int \dfrac{x^2}{\sqrt{9 - 2x^2}} \, dx \)
Solution:
Rewrite denominator as \( \sqrt{9\left(1 - \dfrac{2x^2}{9}\right)} = 3\sqrt{1 - \left(\dfrac{\sqrt{2}}{3}x\right)^2} \).
Let \( \dfrac{\sqrt{2}}{3}x = \sin t \implies x = \dfrac{3}{\sqrt{2}}\sin t \), \( dx = \dfrac{3}{\sqrt{2}}\cos t \, dt \).
Substituting and simplifying yields:
\[ \dfrac{9}{4\sqrt{2}} \left( \arcsin\left(\dfrac{\sqrt{2}}{3}x\right) - \dfrac{1}{2}\sin\left(2\arcsin\left(\dfrac{\sqrt{2}}{3}x\right)\right) \right) + C \]Exercise 2: Evaluate \( \displaystyle \int \dfrac{\sqrt{9x^2 + 1}}{x^4} \, dx \)
Solution:
Let \( 3x = \tan t \implies x = \dfrac{1}{3}\tan t \), \( dx = \dfrac{1}{3}\sec^2 t \, dt \).
Substituting into the integral and simplifying leads to:
\[ -\dfrac{(1 + 9x^2)^{3/2}}{3x^3} + C \]Exercise 3: Evaluate \( \displaystyle \int \dfrac{1}{\sqrt{9x^2 - 36}} \, dx \)
Solution:
Rewrite denominator as \( \sqrt{9(x^2 - 4)} = 3\sqrt{x^2 - 4} \).
Let \( \dfrac{x}{2} = \sec t \implies x = 2\sec t \), \( dx = 2\sec t \tan t \, dt \).
Evaluating the integral yields:
\[ \dfrac{1}{3}\ln\left(\dfrac{\left|\sqrt{x^2 - 4} + x\right|}{2}\right) + C \]