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Integrating tan k x sec j x d x When j Is even

Evaluate tan 6 x sec 4 x d x .

Since the power on sec x is even, rewrite sec 4 x = sec 2 x sec 2 x and use sec 2 x = tan 2 x + 1 to rewrite the first sec 2 x in terms of tan x . Thus,

tan 6 x sec 4 x d x = tan 6 x ( tan 2 x + 1 ) sec 2 x d x Let u = tan x and d u = sec 2 x . = u 6 ( u 2 + 1 ) d u Expand . = ( u 8 + u 6 ) d u Evaluate the integral . = 1 9 u 9 + 1 7 u 7 + C Substitute tan x = u . = 1 9 tan 9 x + 1 7 tan 7 x + C .
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Integrating tan k x sec j x d x When k Is odd

Evaluate tan 5 x sec 3 x d x .

Since the power on tan x is odd, begin by rewriting tan 5 x sec 3 x = tan 4 x sec 2 x sec x tan x . Thus,

tan 5 x sec 3 x = tan 4 x sec 2 x sec x tan x . Write tan 4 x = ( tan 2 x ) 2 . tan 5 x sec 3 x d x = ( tan 2 x ) 2 sec 2 x sec x tan x d x Use tan 2 x = sec 2 x 1. = ( sec 2 x 1 ) 2 sec 2 x sec x tan x d x Let u = sec x and d u = sec x tan x d x . = ( u 2 1 ) 2 u 2 d u Expand . = ( u 6 2 u 4 + u 2 ) d u Integrate . = 1 7 u 7 2 5 u 5 + 1 3 u 3 + C Substitute sec x = u . = 1 7 sec 7 x 2 5 sec 5 x + 1 3 sec 3 x + C .
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Integrating tan k x d x Where k Is odd and k 3

Evaluate tan 3 x d x .

Begin by rewriting tan 3 x = tan x tan 2 x = tan x ( sec 2 x 1 ) = tan x sec 2 x tan x . Thus,

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

For the first integral, use the substitution u = tan x . For the second integral, use the formula.

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Integrating sec 3 x d x

Integrate sec 3 x d x .

This integral requires integration by parts. To begin, let u = sec x and d v = sec 2 x . These choices make d u = sec x tan x and v = tan x . Thus,

sec 3 x d x = sec x tan x tan x sec x tan x d x = sec x tan x tan 2 x sec x d x Simplify . = sec x tan x ( sec 2 x 1 ) sec x d x Substitute tan 2 x = sec 2 x 1. = sec x tan x + sec x d x sec 3 x d x Rewrite . = sec x tan x + ln | sec x + tan x | sec 3 x d x . Evaluate sec x d x .

We now have

sec 3 x d x = sec x tan x + ln | sec x + tan x | sec 3 x d x .

Since the integral sec 3 x d x has reappeared on the right-hand side, we can solve for sec 3 x d x by adding it to both sides. In doing so, we obtain

2 sec 3 x d x = sec x tan x + ln | sec x + tan x | .

Dividing by 2, we arrive at

sec 3 x d x = 1 2 sec x tan x + 1 2 ln | sec x + tan x | + C .
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Evaluate tan 3 x sec 7 x d x .

1 9 sec 9 x 1 7 sec 7 x + C

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Reduction formulas

Evaluating sec n x d x for values of n where n is odd requires integration by parts. In addition, we must also know the value of sec n 2 x d x to evaluate sec n x d x . The evaluation of tan n x d x also requires being able to integrate tan n 2 x d x . To make the process easier, we can derive and apply the following power reduction formulas . These rules allow us to replace the integral of a power of sec x or tan x with the integral of a lower power of sec x or tan x .

Rule: reduction formulas for sec n x d x And tan n x d x

sec n x d x = 1 n 1 sec n 2 x tan x + n 2 n 1 sec n 2 x d x
tan n x d x = 1 n 1 tan n 1 x tan n 2 x d x

The first power reduction rule may be verified by applying integration by parts. The second may be verified by following the strategy outlined for integrating odd powers of tan x .

Revisiting sec 3 x d x

Apply a reduction formula to evaluate sec 3 x d x .

By applying the first reduction formula, we obtain

sec 3 x d x = 1 2 sec x tan x + 1 2 sec x d x = 1 2 sec x tan x + 1 2 ln | sec x + tan x | + C .
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Using a reduction formula

Evaluate tan 4 x d x .

Applying the reduction formula for tan 4 x d x we have

tan 4 x d x = 1 3 tan 3 x tan 2 x d x = 1 3 tan 3 x ( tan x tan 0 x d x ) Apply the reduction formula to tan 2 x d x . = 1 3 tan 3 x tan x + 1 d x Simplify . = 1 3 tan 3 x tan x + x + C . Evaluate 1 d x .
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Apply the reduction formula to sec 5 x d x .

sec 5 x d x = 1 4 sec 3 x tan x 3 4 sec 3 x

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Key concepts

  • Integrals of trigonometric functions can be evaluated by the use of various strategies. These strategies include
    1. Applying trigonometric identities to rewrite the integral so that it may be evaluated by u -substitution
    2. Using integration by parts
    3. Applying trigonometric identities to rewrite products of sines and cosines with different arguments as the sum of individual sine and cosine functions
    4. Applying reduction formulas

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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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