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Wize University Calculus 2 Textbook > Integration Techniques

Trig Sub

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Trigonometric Substitution

Identifying Clues

  • Usually, when we see ...x2...\sqrt{...x^2...}, it's a trig sub question
  • More specifically, it can handle integrals of the form a2−x2, a2+x2, x2−a2\sqrt{a^2-x^2},\ \sqrt{a^2+x^2},\ \sqrt{x^2-a^2}
  • *Sometimes you need to make the integral appera this way. For example
  • Rewrite a2−x2a^2-x^2 as a2−x22\sqrt{a^2-x^2}^2
  • Rewrite x2−4x+7\sqrt{x^2-4x+7} as (x−2)2+3\sqrt{\left(x-2\right)^2+3}

Note--Don't do extra work!

Remember that these forms don't require trig sub:
  • ∫1a2−x2dx=arcsin⁡(xa)+C\int_{ }^{ }\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin\left(\frac{x}{a}\right)+C
  • ∫−1a2−x2dx=arccos⁡(xa)+C\int_{ }^{ }-\frac{1}{\sqrt{a^2-x^2}}dx=\arccos\left(\frac{x}{a}\right)+C
  • ∫1a2+x2dx=1aarctan⁡(xa)+C\int_{ }^{ }\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C

Strategy

ExpressionSubstitutionIntervalTrigonometric Identitya2−x2x=asin⁡θ−π2≤θ≤π2cos⁡2θ=1−sin⁡2θa2+x2x=atan⁡θ−π2<θ<π2sec⁡2θ=tan⁡2θ+1x2−a2x=asec⁡θ0≤θ<π2tan⁡2θ=sec⁡2θ−1or  π≤θ<3π2\def\arraystretch{2.5} \begin{array}{c|c|c|c} Expression & Substitution & Interval &Trigonometric\space Identity \\ \hline \sqrt{a^2-x^2} & x=a\sin\theta & -\frac{\pi}{2}\le\theta\le\frac{\pi}{2} & \cos^2\theta=1-\sin^2\theta\\ \hline \sqrt{a^2+x^2} & x=a\tan\theta & -\frac{\pi}{2}<\theta<\frac{\pi}{2} & \sec^2\theta=\tan^2\theta+1\\ \hline \sqrt{x^2-a^2}& x=a\sec\theta & 0\le\theta<\frac{\pi}{2} & \tan^2\theta=\sec^2\theta-1\\&&or \space \space \pi\le\theta<\frac{3\pi}{2} \end{array}
*aa is the constant; xx is the variable term--can be something like x−1x-1.
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Steps

Example: Trig Sub

Evalue ∫1231x3x2−1dx\displaystyle\int_1^{\frac{2}{\sqrt3}}\frac{1}{x^3\sqrt{x^2-1}}dx

Practice: Trig Sub

Evaluate ∫01x4−x2dx\displaystyle\int_0^1\frac{x}{\sqrt{4-x^2}}dx

*You can technically use a u-substitution for this question, but try to use trig sub for additional practice

Practice Question

Find ∫1x2−2x−3  ⁣dx{\displaystyle\int}\frac{1}{\sqrt{x^2-2x-3}}\de{x}.

Extra Practice
Practice Question: Trig Sub (w/ completing the square)
Find ∫1x2−2x−3  ⁣dx{\displaystyle\int}\frac{1}{\sqrt{x^2-2x-3}}\de{x}.
Practice: Trig Sub
Evaluate ∫01x4−x2dx\displaystyle\int_0^1\frac{x}{\sqrt{4-x^2}}dx
*You can technically use a u-substitution for this question, but try to use trig sub for additional practice
Practice: Trig Sub
Practice: Trig Sub
Evaluate ∫01x4−x2dx\displaystyle\int_0^1\frac{x}{\sqrt{4-x^2}}dx
*You can technically use a u-substitution for this question, but try to use trig sub for additional practice
Practice Question: Trig Sub (w/ completing the square)
Find ∫1x2−2x−3  ⁣dx{\displaystyle\int}\frac{1}{\sqrt{x^2-2x-3}}\de{x}.
Practice: Trig Sub
Practice: Trig Sub
Evaluate ∫01x4−x2dx\displaystyle\int_0^1\frac{x}{\sqrt{4-x^2}}dx
*You can technically use a u-substitution for this question, but try to use trig sub for additional practice
Practice Question: Trig Sub (w/ completing the square)
Practice Question
Find ∫1x2−2x−3  ⁣dx{\displaystyle\int}\frac{1}{\sqrt{x^2-2x-3}}\de{x}.