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  • Solve integration problems involving products and powers of sin x and cos x .
  • Solve integration problems involving products and powers of tan x and sec x .
  • Use reduction formulas to solve trigonometric integrals.

In this section we look at how to integrate a variety of products of trigonometric functions. These integrals are called trigonometric integrals . They are an important part of the integration technique called trigonometric substitution , which is featured in Trigonometric Substitution . This technique allows us to convert algebraic expressions that we may not be able to integrate into expressions involving trigonometric functions, which we may be able to integrate using the techniques described in this section. In addition, these types of integrals appear frequently when we study polar, cylindrical, and spherical coordinate systems later. Let’s begin our study with products of sin x and cos x .

Integrating products and powers of sin x And cos x

A key idea behind the strategy used to integrate combinations of products and powers of sin x and cos x involves rewriting these expressions as sums and differences of integrals of the form sin j x cos x d x or cos j x sin x d x . After rewriting these integrals, we evaluate them using u -substitution. Before describing the general process in detail, let’s take a look at the following examples.

Integrating cos j x sin x d x

Evaluate cos 3 x sin x d x .

Use u -substitution and let u = cos x . In this case, d u = sin x d x . Thus,

cos 3 x sin x d x = u 3 d u = 1 4 u 4 + C = 1 4 cos 4 x + C .
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Evaluate sin 4 x cos x d x .

1 5 sin 5 x + C

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A preliminary example: integrating cos j x sin k x d x Where k Is odd

Evaluate cos 2 x sin 3 x d x .

To convert this integral to integrals of the form cos j x sin x d x , rewrite sin 3 x = sin 2 x sin x and make the substitution sin 2 x = 1 cos 2 x . Thus,

cos 2 x sin 3 x d x = cos 2 x ( 1 cos 2 x ) sin x d x Let u = cos x ; then d u = sin x d x . = u 2 ( 1 u 2 ) d u = ( u 4 u 2 ) d u = 1 5 u 5 1 3 u 3 + C = 1 5 cos 5 x 1 3 cos 3 x + C .
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Evaluate cos 3 x sin 2 x d x .

1 3 sin 3 x 1 5 sin 5 x + C

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In the next example, we see the strategy that must be applied when there are only even powers of sin x and cos x . For integrals of this type, the identities

sin 2 x = 1 2 1 2 cos ( 2 x ) = 1 cos ( 2 x ) 2

and

cos 2 x = 1 2 + 1 2 cos ( 2 x ) = 1 + cos ( 2 x ) 2

are invaluable. These identities are sometimes known as power-reducing identities and they may be derived from the double-angle identity cos ( 2 x ) = cos 2 x sin 2 x and the Pythagorean identity cos 2 x + sin 2 x = 1 .

Integrating an even power of sin x

Evaluate sin 2 x d x .

To evaluate this integral, let’s use the trigonometric identity sin 2 x = 1 2 1 2 cos ( 2 x ) . Thus,

sin 2 x d x = ( 1 2 1 2 cos ( 2 x ) ) d x = 1 2 x 1 4 sin ( 2 x ) + C .
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Evaluate cos 2 x d x .

1 2 x + 1 4 sin ( 2 x ) + C

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The general process for integrating products of powers of sin x and cos x is summarized in the following set of guidelines.

Problem-solving strategy: integrating products and powers of sin x And cos x

To integrate cos j x sin k x d x use the following strategies:

  1. If k is odd, rewrite sin k x = sin k 1 x sin x and use the identity sin 2 x = 1 cos 2 x to rewrite sin k 1 x in terms of cos x . Integrate using the substitution u = cos x . This substitution makes d u = sin x d x .
  2. If j is odd, rewrite cos j x = cos j 1 x cos x and use the identity cos 2 x = 1 sin 2 x to rewrite cos j 1 x in terms of sin x . Integrate using the substitution u = sin x . This substitution makes d u = cos x d x . ( Note : If both j and k are odd, either strategy 1 or strategy 2 may be used.)
  3. If both j and k are even, use sin 2 x = ( 1 / 2 ) ( 1 / 2 ) cos ( 2 x ) and cos 2 x = ( 1 / 2 ) + ( 1 / 2 ) cos ( 2 x ) . After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate.

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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