Lecture in Integration of the Powers of the Trigonometric Functions

In this topic we are going to learn how to integrate certain combinations of trigonometric functions. We will be using some techniques to help us make it easy for us to evaluate each of them. First let's start off with the review of the basic formula of Integration by Substitution, I know at this point you could perfectly evaluate integrals using this method.

Formula:

, where n ≠ –1
, where n = –1

Integration of the powers of sine and cosine functions

• Case 1. Has the form ∫sinⁿax cosax dx. I believe this is the basic and the easiest to evaluate because of the presence of the cosine.

Examples 1: ∫sin⁵3x cos3x dx


Solution: Let u = sin3x, du = 3cos3x dx

=



• Case 2. Has the form ∫cosⁿax sinax dx. In this case instead sine, the one with the exponent is the cosine and it is a lot easier to evaluate because of the presence of sine.

Examples 2: ∫cos³4x sin4x dx


Solution: Let u = cos4x, du = -4sin4x dx

=



• Case 3. Has the form ∫sin^max cosⁿax dx where m is odd integer. The first step is to factor sin x, then change the remaining factor into sin²ax = 1 - cos²ax.

Examples 3:


∫sin³3x dx = ∫sin²3x ∙ sin3x dx
= ∫(1 - cos²3x) sin3x dx
= ∫sin3x dx - ∫cos²3x sin3x dx
=


Examples 4:


∫sin⁵2x cos²2x dx = ∫sin⁴2x ∙ sin2x cos²2x dx
= ∫(sin²2x)² cos²2x sin2x dx
= ∫(1 - cos²2x)² cos²2x sin2x dx
= ∫(1 - 2cos²2x + cos⁴2x) cos²2x sin2x dx
= ∫cos²2x sin2x dx - 2∫cos⁴2x sin2x dx + ∫cos⁶2x sin2x dx

Let u = cos2x, du = -2sin2x dx

=



• Case 4. Has the form ∫sin^max cosⁿax dx where n is odd integer. Here we can factor cos x, then change the remaining factor into cos²ax = 1 - sin²ax.

Examples 5:


∫cos³4x dx = ∫cos²4x ∙ cos4x dx
= ∫(1 - sin²4x) cos4x dx
= ∫cos4x dx - ∫sin²4x cos4x dx
=


Examples 6:


∫sin²x cos³x dx = ∫sin²x cos²x ∙ cosx dx
= ∫sin²x (1 - sin²x) cosx dx
= ∫(sin²x - sin⁴x) cosx dx

Let u = sinx, du = cosx dx

=



• Case 5. Has the form ∫sin^max cosⁿax dx where m and n are even integers. In this case we will be using the half-angle identities below.


and


Examples 7:


∫sin²x dx = ½∫(1 - cos2x)dx
= ½∫dx - ½∫cos2x dx
=


Examples 8:


∫sin²x cos²x dx = ∫½(1 - cos2x)½(1 + cos2x) dx
= ¼∫(1 - cos²2x) dx
= ¼∫[1 - ½(1 + cos4x)] dx
= ⅛∫(1 - cos4x) dx
=



• Case 6. Has the form ∫sinaxcosbx dx, ∫sinaxsinbx dx, ∫cosaxcosbx dx. To find the integrals for this kind of combination of trigonometric functions, we will be needing the identities below.







Examples 9:


∫sin6xcos3x dx = ½∫(sin9x + sin3x) dx
= ½∫sin9x dx + ½∫sin3x dx
=


Examples 10:


∫sin5xsin2x dx = ½∫(cos3x - cos7x) dx
= ½∫cos3x dx - ½∫cos7x dx
=

Integration of the powers of tangent and secant functions

• Case 1. Has the form ∫tan^mu secⁿu du. In this case, when m is odd factor out sec u tan u du and change the remaining tangent into secant using the identity, tan²u = sec²u - 1.

Examples 11:


∫tan³x sec³x dx = ∫tan²x sec²x ∙ secx tanx dx
= ∫(sec²x - 1)sec²x ∙ secx tanx dx
= ∫(sec⁴x - sec²x)secx tanx dx

Let u = secx, du = secx tanx dx

= ∫ u⁴ du - u² du
=


• Case 2. Has the form ∫tan^mu secⁿu du. When n is even greater than 2, factor out sec²u du and replace the remaining secant by tangent using the identity, sec²u = 1 + tan²u.

Examples 12:


∫tan⁵x sec⁴x dx = ∫tan⁵x sec²x ∙ sec²x dx
= ∫tan⁵x(1 + tan²x)sec²x dx
= ∫tan⁵x sec²x dx + ∫tan⁷x sec²x dx

Let u = tanx, du = sec²x dx

= ∫ u⁵ du + u⁷ du
=


• Case 3. Has the form ∫tan^mu secⁿu du. When m is even the integrand is tangent only use the identity, tan²u = sec²u - 1.

Examples 13:


∫tan⁴x dx = ∫tan²x ∙(sec²x - 1) dx
= ∫tan²x sec²x dx - ∫tan²x dx
= ∫tan²x sec²x dx - ∫(sec²x - 1) dx

Let u = tanx, du = sec²x dx

= ∫u² du - ∫du + ∫dx
=

Integration of the powers of cotangent and Cosecant functions

• Case 1. Has the form ∫cot^mu cscⁿu du. When m is odd, factor out csc u cot u du and change the remaining cotangent into cosecant using the identity, cot²u = csc²u - 1.

Examples 14:


∫cot³2x csc³2x dx = ∫cot²2x csc²2x ∙ (cot2x csc2x dx)
= ∫(csc²2x - 1)csc²2x ∙ cot2x csc2x dx
= ∫(csc⁴2x - csc²2x)cot2x csc2x dx

Let u = csc2x, du = -2csc2x cot2x dx

= -½∫u⁴ du + ½∫u² du
=


• Case 2. Has the form ∫cot^mu cscⁿu du. When n is even greater than 2, factor out csc²u du and replace the remaining cosecant by cotangent using the identity, csc²u = 1 + cot²u.

Examples 15:


∫cot⁴x csc⁴x dx = ∫cot⁴x csc²x ∙ csc²x dx
= ∫cot⁴x(1 + cot²x) csc²x dx
= ∫(cot⁴x + cot⁶x) csc²x dx

Let u = cotx, du = -csc²x dx

= -∫u⁴ du - ∫u⁶ du
=

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