Trigonometric Integrals

7 Techniques of Integration

7.2 Trigonometric Integrals

• In this section, we utilize trigonometric identities to integrate specific combinations of trigonometric functions.

Integrals of Powers of Sine and Cosine

Overview

• We begin by analyzing integrals where the integrand consists of a power of sine, a power of cosine, or a product of these trigonometric functions.

Example 2: Solution

• To tackle this integral, we decide against converting the expression entirely into terms of sine due to the absence of an extra cosine factor.

• Instead, we isolate one sine factor and express the remaining factor in terms of cosine:

  • Letting the integrand be expressed as:
    ext{sin}^3 x ext{cos}^2 x ext{dx} = ext{(sin}^2 x)^2 ext{cos}^2 x ext{sin} x ext{dx}

  • This can be rewritten using the identity ext{(1 - cos}^2 x) leading to:
    ext{(1 - u}^2)^2 u^2 (-du) where we set u = ext{cos} x and du = - ext{sin} x ext{dx}.

  • Expanding and simplifying gives:
    - rac{1}{3}u^3 + rac{1}{2}u^2 + C

  • In terms of ext{cos} x this is:
    - rac{1}{3} ext{cos}^3 x + rac{1}{2} ext{cos}^2 x + C

Example 3: Solution

• When faced with the integral, we find no easier means to evaluate it directly. Instead, applying the half-angle formula leads to simplifying approaches:

  • By integrating using the transformation u = 2x when handling ext{cos} 2x, we derive:
    rac{1}{12} = rac{1}{2} ext{sin}^2( rac{ ext{π}}{2})

  • Here, careful substitution helped in the resolution of the integral.

Guidelines for Evaluating Integrals of Powers of Sine and Cosine

• The following strategies summarize approaches when working with integrals of the forms:

  • If the power of cosine is odd (n = 2k + 1):

  • Save one cosine factor; use it to express remaining factors in terms of sine.

  • Substitute: u = ext{sin} x.

  • If the power of sine is odd (m = 2k + 1):

  • Save one sine factor; use it to re-express the rest in terms of cosine.

  • Substitute: u = ext{cos} x.

  • If both powers are even:

  • Apply half-angle identities within the integral.

Integrals of Powers of Secant and Tangent

Overview

• For integrals of secant and tangent, a similar strategy can be adopted:

  • Isolate a factor, converting the even power of secant to a tangent expression where beneficial, or the other way around.

Example 5: Solution

• We isolate a secant factor, expressing the remaining factor in terms of tangent, for evaluation:

  • Utilizing the relation between tangent and secant leads us to:

  • ext{sec}^n x ext{tan}^m x ext{dx}

  • Transform this into integral forms of u by substitution: u = ext{tan} x.

  • Final evaluation yields:
    rac{1}{2} ext{tan} x + rac{1}{6} ext{tan}^3 x + C

Strategies for Secant and Tangent

• When evaluating:

  • If the secant power is even (n = 2k, k ≥ 2):

  • Save a sec x factor to express remaining factors in terms of tangent. Make the substitution u = ext{tan} x.

  • If the tangent power is odd (m = 2k + 1):

  • Save a sec x tan x factor while expressing the other terms as secant powers leading to substitution: u = ext{sec} x.

  • For other configurations: Various identities and methods such as integration by parts may be employed for solutions.

Important Integral Formulas

• The integral of tangent can be expressed as:

  • ext{tan} x ext{dx} = ext{ln}| ext{sec} x| + C

• Fundamental operation involving secant:

  • To establish this, the numerator and denominator can be multiplied by ext{sec} x + ext{tan} x, simplifying subsequent integrations.

Example 7: Find Solution

• The integral only consists of tangent; we opt to write a factor in terms of:

  • rac{1}{2} - ext{tan}^2 x - ext{ln}| ext{sec} x| + C

• Employing the substitution u = ext{tan} x allows implicit transition between trigonometrical expressions and integrals.

Using Product Identities

Overview

• Product identities prove beneficial for calculating several trigonometric integrals. For computations involving:

  • (a)

  • (b)

  • (c)

  • The respective identities are required to efficiently solve these integrals.

Example 9: Evaluate Solution

• Using integration by parts is an option, yet leveraging product identities often provides a more straightforward solution. For example:

  • Applying specific product identities leads to simplified evaluation of the integral.