\( \newcommand{\sech}{\operatorname{sech}} \) \( \newcommand{\inverse}[1]{#1^\leftarrow} \) \( \newcommand{\<}{\langle} \) \( \newcommand{\>}{\rangle} \) \( \newcommand{\vect}{\mathbf} \) \( \newcommand{\veci}{\mathbf{\hat ı}} \) \( \newcommand{\vecj}{\mathbf{\hat ȷ}} \) \( \newcommand{\veck}{\mathbf{\hat k}} \) \( \newcommand{\curl}{\operatorname{curl}\,} \) \( \newcommand{\dv}{\operatorname{div}\,} \) \( \newcommand{\detThree}[9]{ \operatorname{det}\left( \begin{array}{c c c} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9 \end{array} \right) } \) \( \newcommand{\detTwo}[4]{ \operatorname{det}\left( \begin{array}{c c} #1 & #2 \\ #3 & #4 \end{array} \right) } \)

Section 2.3 Calculus 2

Trigonometric Substitution
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2.3 Trigonometric Substitution

2.3.1 Substituting for \(a+bx^2\)

  • To eliminate factors of the form \(a+bx^2\) from an integral, use the substitution \(a+bx^2=a+a\tan^2\theta=a\sec^2\theta\) with \(-\pi/2<\theta<\pi/2\).
  • Example Find \(\int\frac{z^2}{4+9z^2}\,dz\).
  • Example Compute \(\int_0^2\frac{1}{\sqrt{16+4x^2}}\,dx\). (Recall \(\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+C\).)

2.3.2 Substituting for \(a-bx^2\)

  • To eliminate factors of the form \(a-bx^2\) from an integral, use the substitution \(a-bx^2=a-a\sin^2\theta=a\cos^2\theta\) with \(-\pi/2\leq\theta\leq\pi/2\).
  • Note that this is only valid when \(|x|\leq\sqrt{a/b}\), which is guaranteed when \(a-bx^2\) is under a square root.
  • Example Find \(\int(4-25s^2)^{-3/2}\,ds\).
  • Example Find \(\int\frac{x^3}{\sqrt{1-4x^2}}\,dx\).

2.3.3 Substituting for \(bx^2-a\)

  • To eliminate factors of the form \(bx^2-a\) from an integral, use the substitution \(bx^2-a=a\sec^2\theta-a=a\tan^2\theta\) with \(0\leq\theta<\pi/2\).
  • Note that this is only valid when \(x\geq\sqrt{a/b}\), which will be assumed in our problems.
  • Example Prove \(\int\frac{1}{x\sqrt{x^2-1}}\,dx=\inverse\sec x+C\) where \(x>1\).
  • Example Find \(\int\frac{\sqrt{y^2-16}}{y}\,dy\) where \(y\geq 4\).

2.3.4 Using Inverse Trigonometric Antiderivatives

  • Sometimes, a simpler substiution may be combined with the following antiderivatives to obtain a solution more elegantly.
    • \(\int\frac{1}{1+x^2}\,dx=\inverse\tan x+C\).
    • \(\int\frac{1}{\sqrt{1-x^2}}\,dx=\inverse\sin x + C\).
    • \(\int\frac{1}{x\sqrt{x^2-1}}\,dx=\inverse\sec x+C\) where \(x>1\).
  • Example Find \(\int\frac{3}{\sqrt{9-x^2}}\,dx\) without using a trigonometric substitution.

Review Exercises

  1. Find \(\int\frac{2}{\sqrt{1+4z^2}}\,dz\).
  2. Find \(\int\frac{x^3}{9+x^2}\,dx\).
  3. Find \(\int \frac{4}{(1-y^2)^{3/2}}\,dy\).
  4. Find \(\int\frac{2x^3}{\sqrt{9-x^2}}\,dx\).
  5. Prove \(\int\frac{1}{\sqrt{1-x^2}}\,dx=\inverse\sin x+C\).
  6. Find \(\int\frac{\sqrt{x^2-16}}{x}\,dx\) where \(x\geq 4\).
  7. Find \(\int\frac{1}{\sqrt{4t^2-1}}\,dt\) where \(t>\frac{1}{2}\).
  8. Find \(\int\frac{2}{\sqrt{1-4x^2}}\,dx\) without a trigonometric substitution.
  9. Find \(\int\frac{2}{4+9x^2}\,dx\) without a trigonometric substitution.
  10. Find \(\int \frac{1}{\sqrt{9+y^2}}\,dy\).
  11. Find \(\int \frac{1}{x\sqrt{4x^2-1}}\,dy\) where \(x>\frac{1}{2}\).

Solutions 1-5

Solutions 6-11

Textbook References

  • University Calculus: Early Transcendentals (3rd Ed)
    • 8.3