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Section 5.3 Integration of Trigonometry (TI3)

Subsection 5.3.1 Activities

Activity 5.3.1.

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

Activity 5.3.2.

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

Activity 5.3.3.

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

Activity 5.3.4.

It’s possible to use substitution to evaluate ∫sin4⁑(x)cos3⁑(x)dx, by taking advantage of the trigonometric identity sin2⁑(x)+cos2⁑(x)=1.
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)(1βˆ’?)cos⁑(x)dx=∫?(1βˆ’?)du=∫(u4βˆ’u6)du=15u5βˆ’17u7+C=?

Activity 5.3.5.

Trying to substitute u=cos⁑(x),du=βˆ’sin⁑(x)dx in the previous example is less successful.
∫sin4⁑(x)cos3⁑(x)dx=βˆ’βˆ«sin3⁑(x)cos3⁑(x)(βˆ’sin⁑(x)dx)=βˆ’βˆ«sin3⁑(x)u3du=β‹―?
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)

Activity 5.3.6.

Try to show
∫sin5⁑(x)cos2⁑(x)dx=βˆ’17cos7⁑(x)+25cos5⁑(x)βˆ’13cos3⁑(x)+C
by first trying u=sin⁑(x), and then trying u=cos⁑(x) instead.
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.

Observation 5.3.7.

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).

Activity 5.3.8.

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

Activity 5.3.9.

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

Activity 5.3.10.

Consider ∫sin4⁑(x)cos4⁑(x)dx. Which would be the most useful way to rewrite the integral?
  1. ∫(1βˆ’cos2⁑(x))2cos4⁑(x)dx
  2. ∫sin4⁑(x)(1βˆ’sin2⁑(x))2dx
  3. ∫(1βˆ’cos⁑(2x)2)2(1+cos⁑(2x)2)2dx

Activity 5.3.11.

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

Remark 5.3.12.

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

Subsection 5.3.2 Videos

Figure 108. Video: Compute integrals involving products of trigonometric functions

Subsection 5.3.3 Exercises