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Calculus for Team-Based Inquiry Learning: 2024 Early Edition

TBIL Institute Fellows

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Section 5.3 Integration of trigonometry (TI3)

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Subsection 5.3.1 Activities

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Activity 5.3.1.

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Consider ∫sin⁡(x)cos⁡(x)dx. Which substitution would you choose to evaluate this integral?
  1. u=sin⁡(x)
  2. u=cos⁡(x)
  3. u=sin⁡(x)cos⁡(x)
  4. Substitution is not effective
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Activity 5.3.2.

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Consider ∫sin4⁡(x)cos⁡(x)dx. Which substitution would you choose to evaluate this integral?
  1. u=sin⁡(x)
  2. u=sin4⁡(x)
  3. u=cos⁡(x)
  4. Substitution is not effective
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Activity 5.3.3.

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Consider ∫sin4⁡(x)cos3⁡(x)dx. Which substitution would you choose to evaluate this integral?
  1. u=sin⁡(x)
  2. u=cos3⁡(x)
  3. u=cos⁡(x)
  4. Substitution is not effective
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Activity 5.3.4.

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It’s possible to use subtitution to evaluate ∫sin4⁡(x)cos3⁡(x)dx, by taking advantage of the trigonometric identity sin2⁡(x)+cos2⁡(x)=1.
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Complete the following substitution of u=sin⁡(x),du=cos⁡(x)dx by filling in the missing ?s.
∫sin4⁡(x)cos3⁡(x)dx=∫sin4⁡(x)(?)cos⁡(x)dx=∫sin4⁡(x)(1−?)cos⁡(x)dx=∫?(1−?)du=∫(u4−u6)du=15u5−17u7+C=?
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Activity 5.3.5.

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Trying to substitute u=cos⁡(x),du=−sin⁡(x)dx in the previous example is less successful.
∫sin4⁡(x)cos3⁡(x)dx=−∫sin3⁡(x)cos3⁡(x)(−sin⁡(x)dx)=−∫sin3⁡(x)u3du=⋯?
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Which feature of sin4⁡(x)cos3⁡(x) made u=sin⁡(x) the better choice?
  1. The even power of sin4⁡(x)
  2. The odd power of cos3⁡(x)
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Activity 5.3.6.

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Try to show
∫sin5⁡(x)cos2⁡(x)dx=−17cos7⁡(x)+25cos5⁡(x)−13cos3⁡(x)+C
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by first trying u=sin⁡(x), and then trying u=cos⁡(x) instead.
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Which substitution worked better and why?
  1. u=sin⁡(x) due to sin5⁡(x)’s odd power.
  2. u=sin⁡(x) due to cos2⁡(x)’s even power.
  3. u=cos⁡(x) due to sin5⁡(x)’s odd power.
  4. u=cos⁡(x) due to cos2⁡(x)’s even power.
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Observation 5.3.7.

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When integrating the form ∫sinm⁡(x)cosn⁡(x)dx:
  • If sin’s power is odd, rewrite the integral as ∫g(cos⁡(x))sin⁡(x)dx and use u=cos⁡(x).
  • If cos’s power is odd, rewrite the integral as ∫h(sin⁡(x))cos⁡(x)dx and use u=sin⁡(x).
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Activity 5.3.8.

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Let’s consider ∫sin2⁡(x)dx.
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(a)
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Use the fact that sin2⁡(θ)=1−cos⁡(2θ)2 to rewrite the integrand using the above identities as an integral involving cos⁡(2x).
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(b)
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Show that the integral evaluates to 12x−14sin⁡(2x)+C.
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Activity 5.3.9.

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Let’s consider ∫sin2⁡(x)cos2⁡(x)dx.
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(a)
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Use the fact that cos2⁡(θ)=1+cos⁡(2θ)2 and sin2⁡(θ)=1−cos⁡(2θ)2 to rewrite the integrand using the above identities as an integral involving cos2⁡(2x).
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(b)
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Use the above identities to rewrite this new integrand as one involving cos⁡(4x).
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(c)
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Show that integral evaluates to 18x−132sin⁡(4x)+C.
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Activity 5.3.10.

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Consider ∫sin4⁡(x)cos4⁡(x)dx. Which would be the most useful way to rewrite the integral?
  1. ∫(1−cos2⁡(x))2cos4⁡(x)dx
  2. ∫sin4⁡(x)(1−sin2⁡(x))2dx
  3. ∫(1−cos⁡(2x)2)2(1+cos⁡(2x)2)2dx
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Activity 5.3.11.

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Consider ∫sin3⁡(x)cos5⁡(x)dx. Which would be the most useful way to rewrite the integral?
  1. ∫(1−cos2⁡(x))cos5⁡(x)sin⁡(x)dx
  2. ∫sin3⁡(x)(1+cos⁡(2x)2)2cos⁡(x)dx
  3. ∫sin3⁡(x)(1−sin2⁡(x))2cos⁡(x)dx
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Remark 5.3.12.

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We might also use some other trigonometric identities to manipulate our integrands, listed in Appendix B.
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Activity 5.3.13.

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Consider ∫sin⁡(θ)sin⁡(3θ)dθ.
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(b)
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Rewrite the integrand using the selected identity.
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Subsection 5.3.2 Videos

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Figure 107. Video: Compute integrals involving products of trigonometric functions
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