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Integrating cos j x sin k x d x Where k Is odd

Evaluate cos 8 x sin 5 x d x .

Since the power on 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 .
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Integrating cos j x sin k x d x Where k And j Are even

Evaluate sin 4 x d x .

Since the power on sin x is even ( k = 4 ) and the power on cos x is even ( 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 .

Since cos 2 ( 2 x ) has an even power, substitute 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 .
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Evaluate cos 3 x d x .

sin x 1 3 sin 3 x + C

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Evaluate cos 2 ( 3 x ) d x .

1 2 x + 1 12 sin ( 6 x ) + C

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In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include sin ( a x ) , sin ( b x ) , cos ( a x ) , and cos ( b x ) . These integrals are evaluated by applying trigonometric identities, as outlined in the following rule.

Rule: integrating products of sines and cosines of different angles

To integrate products involving sin ( a x ) , sin ( b x ) , cos ( a x ) , and cos ( b x ) , use the substitutions

sin ( a x ) sin ( b x ) = 1 2 cos ( ( a b ) x ) 1 2 cos ( ( a + b ) x )
sin ( a x ) cos ( b x ) = 1 2 sin ( ( a b ) x ) + 1 2 sin ( ( a + b ) x )
cos ( a x ) cos ( b x ) = 1 2 cos ( ( a b ) x ) + 1 2 cos ( ( a + b ) x )

These formulas may be derived from the sum-of-angle formulas for sine and cosine.

Evaluating sin ( a x ) cos ( b x ) d x

Evaluate sin ( 5 x ) cos ( 3 x ) d x .

Apply the identity sin ( 5 x ) cos ( 3 x ) = 1 2 sin ( 2 x ) 1 2 cos ( 8 x ) . Thus,

sin ( 5 x ) cos ( 3 x ) d x = 1 2 sin ( 2 x ) 1 2 cos ( 8 x ) d x = 1 4 cos ( 2 x ) 1 16 sin ( 8 x ) + C .
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Evaluate cos ( 6 x ) cos ( 5 x ) d x .

1 2 sin x + 1 22 sin ( 11 x ) + C

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Integrating products and powers of tan x And sec x

Before discussing the integration of products and powers of tan x and sec x , it is useful to recall the integrals involving tan x and sec x we have already learned:

  1. sec 2 x d x = tan x + C
  2. sec x tan x d x = sec x + C
  3. tan x d x = ln | sec x | + C
  4. sec x d x = ln | sec x + tan x | + C .

For most integrals of products and powers of tan x and sec x , we rewrite the expression we wish to integrate as the sum or difference of integrals of the form tan j x sec 2 x d x or sec j x tan x d x . As we see in the following example, we can evaluate these new integrals by using u -substitution.

Evaluating sec j x tan x d x

Evaluate sec 5 x tan x d x .

Start by rewriting sec 5 x tan x as sec 4 x sec x tan x .

sec 5 x tan x d x = sec 4 x sec x tan x d x Let u = sec x ; then , d u = sec x tan x d x . = u 4 d u Evaluate the integral . = 1 5 u 5 + C Substitute sec x = u . = 1 5 sec 5 x + C
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You can read some interesting information at this website to learn about a common integral involving the secant.

Evaluate tan 5 x sec 2 x d x .

1 6 tan 6 x + C

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We now take a look at the various strategies for integrating products and powers of sec x and tan x .

Problem-solving strategy: integrating tan k x sec j x d x

To integrate tan k x sec j x d x , use the following strategies:

  1. If j is even and j 2 , rewrite sec j x = sec j 2 x sec 2 x and use sec 2 x = tan 2 x + 1 to rewrite sc j 2 x in terms of tan x . Let u = tan x and d u = sec 2 x .
  2. If k is odd and j 1 , rewrite tan k x sec j x = tan k 1 x sec j 1 x sec x tan x and use tan 2 x = sec 2 x 1 to rewrite tan k 1 x in terms of sec x . Let u = sec x and d u = sec x tan x d x . ( Note : If j is even and k is odd, then either strategy 1 or strategy 2 may be used.)
  3. If k is odd where k 3 and j = 0 , rewrite tan k x = tan k 2 x tan 2 x = tan k 2 x ( sec 2 x 1 ) = tan k 2 x sec 2 x tan k 2 x . It may be necessary to repeat this process on the tan k 2 x term.
  4. If k is even and j is odd, then use tan 2 x = sec 2 x 1 to express tan k x in terms of sec x . Use integration by parts to integrate odd powers of sec x .

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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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