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Calculus for Team-Based Inquiry Learning: 2023 Early Edition

TBIL Institute Fellows

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Section 5.4 Trigonometric Substitution (TI4)

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Subsection 5.4.1 Activities

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Activity 5.4.1.

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Consider ∫9−4x2dx. Which substitution would you choose to evaluate this integral?
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    u=9−4x2
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    u=9−4x2
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    u=3−2x
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    Substitution is not effective
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Activity 5.4.2.

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To find ∫9−4x2dx, we will need a more advanced substitution. Which of these candidates is most reasonable?
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    Let v satisfy 9−4x2=9−9e2v=9e−2v.
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    Let θ satisfy 9−4x2=9−9sin2⁡θ=9cos2⁡θ.
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    Let w satisfy 9−4x2=4−8ln⁡|w|=4ln⁡|2w|.
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    Let ϕ satisfy 9−4x2=4−4cos2⁡ϕ=4sin2⁡ϕ.
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Activity 5.4.3.

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Fill in the missing ?s for the following calculation.
Let 9−4x2=9−9sin2⁡θ=9cos2⁡θ4x2=?x=?dx=?dθ
∫9−4x2dx=∫?(?dθ)=∫92cos2⁡θdθ
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Activity 5.4.4.

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From Section 5.3 we may find ∫cos2⁡θdθ=12θ+12sin⁡θcos⁡θ+C.
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Use this to continue your work in the previous activity and complete the integration by trigonometric substitution.
sin⁡(θ)=?θ=arcsin⁡(?)cos⁡(θ)=??
∫9−4x2dx=⋯=∫92cos2⁡θdθ=92(12θ+12sin⁡θcos⁡θ)+C=94(?)+94(?)(?)+C
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Activity 5.4.5.

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Use similar reasoning to complete the following proof that ddx[arcsin⁡(x)]=11−x2.
Let 1−x2=1−?θ=?θx2=?x=?dx=?dθθ=?
∫11−x2dx=∫1?(?dθ)=∫dθ=?+C=arcsin⁡(x)+C
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Activity 5.4.6.

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Substitutions of the form
16−25x2=16−16sin2⁡x=16cos2⁡x
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are made possible due to the Pythagorean identity sin2⁡(x)+cos2⁡(x)=1.
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Which two of these four identities can be obtained from dividing both sides of sin2⁡(x)+cos2⁡(x)=1 by cos2⁡(x) and rearranging?
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    tan2⁡(x)−1=sec2⁡(x)
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    tan2⁡(x)+1=sec2⁡(x)
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    sec2⁡(x)−1=tan2⁡(x)
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    sec2⁡(x)+1=tan2⁡(x)
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Observation 5.4.7.

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In summary, certain quadratic expressions inside an integral may be substituted with trigonometric functions to take advantage of trigonometric identities and simplify the integrand:
Let b−ax2=b−bsin2⁡(θ)=bcos2⁡(θ)So x=basin⁡(θ)
Let b+ax2=b+btan2⁡(θ)=bsec2⁡(θ)So x=batan⁡(θ)
Let ax2−b=bsec2⁡(θ)−b=btan2⁡(θ)So x=basec⁡(θ)
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Activity 5.4.8.

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Complete the following trignometric substitution to find ∫34+25x2dx.
Let 4+25x2=2+?θ=?θ25x2=?x=?dx=?dθθ=?
∫34+25x2dx=∫3?(?dθ)=∫?dθ=?+C=310arctan⁡(52x)+C
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Activity 5.4.9.

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Complete the following trignometric substitution to find ∫7x9x2−16dx.
Let 9x2−16=?θ−16=?θ9x2=?x=?dx=?dθθ=?
∫7x9x2−16dx=∫7??(?dθ)=∫?dθ=?+C=74arcsec(34x)+C
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Activity 5.4.10.

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Use appropriate trignometric substitutions and the given trigonometric integrals to find each of the following.
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(a)
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∫−9x2+16x2dx=⋯=∫3cos2⁡θsin2⁡θdθ=−3θ−3cos⁡θsin⁡θ+C=−3arcsin⁡(?)−??+C
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(b)
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∫29x2−16xdx=⋯=∫8tan2⁡θdθ=8tan⁡θ−8θ+C=??−8arcsec(?)+C
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(c)
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∫181x2+4dx=⋯=∫19sec⁡θdθ=19log⁡|sec⁡θ+tan⁡θ|+C=19log⁡|?+12?|+C
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Activity 5.4.11.

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Consider the unit circle x2+y2=1. Find a function f(x) so that y=f(x) is the graph of the upper-half semicircle of the unit circle.
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Subsection 5.4.2 Videos

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Figure 108. Video: Use trigonometric substitution to compute indefinite integrals
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