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 sin x and cos x . cos x .
A key idea behind the strategy used to integrate combinations of products and powers of sin x sin x and cos x cos x involves rewriting these expressions as sums and differences of integrals of the form ∫ sin j x cos x d x ∫ sin j x cos x d x or ∫ cos j x sin x d x . ∫ 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.
Evaluate ∫ cos 3 x sin x d x . ∫ cos 3 x sin x d x .
Use u u -substitution and let u = cos x . u = cos x . In this case, d u = − sin x d x . 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 . ∫ cos 3 x sin x d x = − ∫ u 3 d u = − 1 4 u 4 + C = − 1 4 cos 4 x + C .
Evaluate ∫ sin 4 x cos x d x . ∫ sin 4 x cos x d x .
Evaluate ∫ cos 2 x sin 3 x d x . ∫ cos 2 x sin 3 x d x .
To convert this integral to integrals of the form ∫ cos j x sin x d x , ∫ cos j x sin x d x , rewrite sin 3 x = sin 2 x sin x sin 3 x = sin 2 x sin x and make the substitution sin 2 x = 1 − cos 2 x . 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 . ∫ 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 .
Evaluate ∫ cos 3 x sin 2 x d x . ∫ cos 3 x sin 2 x d x .
In the next example, we see the strategy that must be applied when there are only even powers of sin x sin x and cos x . cos x . For integrals of this type, the identities
sin 2 x = 1 2 − 1 2 cos ( 2 x ) = 1 − cos ( 2 x ) 2 sin 2 x = 1 2 − 1 2 cos ( 2 x ) = 1 − cos ( 2 x ) 2
cos 2 x = 1 2 + 1 2 cos ( 2 x ) = 1 + cos ( 2 x ) 2 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 cos ( 2 x ) = cos 2 x − sin 2 x and the Pythagorean identity cos 2 x + sin 2 x = 1 . cos 2 x + sin 2 x = 1 .
Evaluate ∫ sin 2 x d x . ∫ 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 ) . 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 . ∫ sin 2 x d x = ∫ ( 1 2 − 1 2 cos ( 2 x ) ) d x = 1 2 x − 1 4 sin ( 2 x ) + C .
Evaluate ∫ cos 2 x d x . ∫ cos 2 x d x .
The general process for integrating products of powers of sin x sin x and cos x cos x is summarized in the following set of guidelines.
To integrate ∫ cos j x sin k x d x ∫ cos j x sin k x d x use the following strategies:
Evaluate ∫ cos 8 x sin 5 x d x . ∫ cos 8 x sin 5 x d x .
Since the power on sin x sin x is odd, use strategy 1. Thus,
∫ cos 8 x sin 5 x d x = ∫ cos 8 x sin 4 x sin x d x Break off sin x . = ∫ cos 8 x ( sin 2 x ) 2 sin x d x Rewrite sin 4 x = ( sin 2 x ) 2 . = ∫ cos 8 x ( 1 − cos 2 x ) 2 sin x d x Substitute sin 2 x = 1 − cos 2 x . = ∫ u 8 ( 1 − u 2 ) 2 ( − d u ) Let u = cos x and d u = − sin x d x . = ∫ ( − u 8 + 2 u 10 − u 12 ) d u Expand . = − 1 9 u 9 + 2 11 u 11 − 1 13 u 13 + C Evaluate the integral . = − 1 9 cos 9 x + 2 11 cos 11 x − 1 13 cos 13 x + C . Substitute u = cos x . ∫ cos 8 x sin 5 x d x = ∫ cos 8 x sin 4 x sin x d x Break off sin x . = ∫ cos 8 x ( sin 2 x ) 2 sin x d x Rewrite sin 4 x = ( sin 2 x ) 2 . = ∫ cos 8 x ( 1 − cos 2 x ) 2 sin x d x Substitute sin 2 x = 1 − cos 2 x . = ∫ u 8 ( 1 − u 2 ) 2 ( − d u ) Let u = cos x and d u = − sin x d x . = ∫ ( − u 8 + 2 u 10 − u 12 ) d u Expand . = − 1 9 u 9 + 2 11 u 11 − 1 13 u 13 + C Evaluate the integral . = − 1 9 cos 9 x + 2 11 cos 11 x − 1 13 cos 13 x + C . Substitute u = cos x .
Evaluate ∫ sin 4 x d x . ∫ sin 4 x d x .
Since the power on sin x sin x is even ( k = 4 ) ( k = 4 ) and the power on cos x cos x is even ( j = 0 ) , ( j = 0 ) , we must use strategy 3. Thus,
∫ sin 4 x d x = ∫ ( sin 2 x ) 2 d x Rewrite sin 4 x = ( sin 2 x ) 2 . = ∫ ( 1 2 − 1 2 cos ( 2 x ) ) 2 d x Substitute sin 2 x = 1 2 − 1 2 cos ( 2 x ) . = ∫ ( 1 4 − 1 2 cos ( 2 x ) + 1 4 cos 2 ( 2 x ) ) d x Expand ( 1 2 − 1 2 cos ( 2 x ) ) 2 . = ∫ ( 1 4 − 1 2 cos ( 2 x ) + 1 4 ( 1 2 + 1 2 cos ( 4 x ) ) d x . ∫ sin 4 x d x = ∫ ( sin 2 x ) 2 d x Rewrite sin 4 x = ( sin 2 x ) 2 . = ∫ ( 1 2 − 1 2 cos ( 2 x ) ) 2 d x Substitute sin 2 x = 1 2 − 1 2 cos ( 2 x ) . = ∫ ( 1 4 − 1 2 cos ( 2 x ) + 1 4 cos 2 ( 2 x ) ) d x Expand ( 1 2 − 1 2 cos ( 2 x ) ) 2 . = ∫ ( 1 4 − 1 2 cos ( 2 x ) + 1 4 ( 1 2 + 1 2 cos ( 4 x ) ) d x .
Since cos 2 ( 2 x ) cos 2 ( 2 x ) has an even power, substitute cos 2 ( 2 x ) = 1 2 + 1 2 cos ( 4 x ) : cos 2 ( 2 x ) = 1 2 + 1 2 cos ( 4 x ) :
= ∫ ( 3 8 − 1 2 cos ( 2 x ) + 1 8 cos ( 4 x ) ) d x Simplify . = 3 8 x − 1 4 sin ( 2 x ) + 1 32 sin ( 4 x ) + C Evaluate the integral . = ∫ ( 3 8 − 1 2 cos ( 2 x ) + 1 8 cos ( 4 x ) ) d x Simplify . = 3 8 x − 1 4 sin ( 2 x ) + 1 32 sin ( 4 x ) + C Evaluate the integral .