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Mathematics LibreTexts

2.2: Trigonometric Substitution

[ "article:topic", "trigonometric substitution", "authorname:green", "showtoc:no" ]
  • Page ID
    522
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    When we have integrals that involve the square root term

    \[\sqrt{a^2+x^2} \]

    we may be able to trigonometric substitution to solve the integral.

    Example \(\PageIndex{1}\)

    Solve

    \[\int \sqrt{1-x^2}dx\]

    by substituting \(x=\sin \theta\) and \(dx=\cos \theta \, d\theta \).

    The integrand then becomes

    \[\begin{align}  \sqrt{1-x^2} &= \sqrt{1-\sin^2 \theta} \\ &=  \sqrt{\cos^2 \theta} \\ &= \cos \theta   \end{align}\]

    We have

    \[\begin{align} \int \sqrt{1-x^2}\; dx &= \int \cos\theta\cos\theta \;d\theta \\ &= \int \cos^2\theta \;d\theta \\ &= \int \Big( \dfrac{1}{2}+\dfrac{1}{2}\cos 2\theta \Big)\; d\theta \\ &= \dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta +C \\ &= \dfrac{1}{2} \arcsin x +\dfrac{1}{2} \sin\theta\cos\theta+C \\ &= \dfrac{1}{2}\arcsin x +\dfrac{1}{2}x\sqrt{1-x^2}+C \end{align}\]

    Exercises \(\PageIndex{1}\)

    1. \[ \int \dfrac{\sqrt{1-x^2}}{x^4} dx \]
    2. \[ \int \dfrac{1}{\sqrt{4-9x^2}}dx \]

    Two Key Formulas

    From Trigonometry, we have the following two key formulas:

    \[ \sec^2\, x = 1 + \tan^2\, x\]

    so

    \[\sec x = \sqrt{1+\tan^2 x} \]

    and

    \[ \tan^2\, x = \sec^2\, x - 1 \]

    so

    \[ \tan x = \sqrt{\sec^2 \, x -1}.\]

    When we have integrals that involve any of the above square roots, we can use the appropriate substitution.

    Example \(\PageIndex{2}\)

    \[\begin{align}  \int \dfrac{x^3}{\sqrt{1+x^2}}dx \\ x= \tan\theta, \; dx=\sec^2\theta \; d\theta \\ \sqrt{1+x^2}=\sqrt{1+\tan^2\theta}= \sqrt{\sec^2\theta} = \sec\theta \\ &= \int \dfrac{\tan^3\theta}{\sec\theta}\sec^2\theta \; d\theta \\  &= \int \tan^3\theta \sec\theta \; d\theta \\ &= \int \tan^2\theta \tan\theta \sec\theta \; d\theta \\ &= \int (\sec^2\theta-1)\sec\theta\tan\theta \; d\theta \\ u=\sec\theta, \; du=\sec\theta\tan\theta \; d\theta \\ &= \int (u^2-1) \; du \\ &=\dfrac{u^3}{3}-u+C \\ &= \dfrac{\sec^3\theta}{3}-\sec\theta+C \\ &= \dfrac{(1+x^2)^{\frac{3}{2}}}{3}-\sqrt{1+x^2}+C \end{align}\]

    Exercise \(\PageIndex{1}\)

    1. \[ \int \dfrac{x^3}{\sqrt{x^2-1}} \; dx   \]
    2. \[ \int \dfrac{x^2}{\sqrt{9+4x^2}} \; dx  \]
    3. \[ \int \dfrac{1}{\sqrt{x^2+2x}} \; dx  \]

    Larry Green (Lake Tahoe Community College)

    • Integrated by Justin Marshall.