7.2 Trigonometric Integrals

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7.2 Trigonometric Integrals

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Trigonometric Integrals

In this section we use trigonometric identities to integrate certain combinations of trigonometric functions.

We start with powers of sine and cosine.

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Example 2

Find ∫

^sin⁵^{x cos}²^{x dx.}

Solution:

We could convert cos²x to 1 – sin²x, but we would be left with an expression in terms of sin x with no extra cos x factor.

Instead, we separate a single sine factor and rewrite the remaining sin⁴x factor in terms of cos x:

sin⁵x cos²x = (sin²x)² cos²x sin x

= (1 – cos²x)² cos²x sin x

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Example 2 – Solution

Substituting u = cos x, we have du = –sin x dx and so

∫

^sin⁵^{x cos}²^{x dx =}

∫

^(sin²^x)² ^cos²^{x sin x dx}

=

∫

^{(1 – cos}²^x)² ^cos²^{x sin x dx}

=

∫

^{(1 – u}²⁾²^u² ^{(–du) = –}

∫

^(u² ^{– 2u}⁴ ^{+ u}⁶^)du

=

= – cos³x + cos⁵x – cos⁷x + C

cont’d

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Example 3

Evaluate

Solution:

If we write sin²x = 1 – cos²x, the integral is no simpler to evaluate. Using the half-angle formula for sin²x, however, we have

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Example 3 – Solution

Notice that we mentally made the substitution u = 2x when integrating cos 2x.

cont’d

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Trigonometric Integrals

To summarize, we list guidelines to follow when evaluating integrals of the form

∫

^sin^m^{x cos}ⁿx dx, where m ≥ 0 and

n ≥ 0 are integers.

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Trigonometric Integrals

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Trigonometric Integrals

We can use a similar strategy to evaluate integrals of the form

∫

^tan^m^{x sec}ⁿ^{x dx.}

Since (d/dx) tan x = sec²x, we can separate a sec²x factor and convert the remaining (even) power of secant to an expression involving tangent using the identity

sec²x = 1 + tan²x.

Or, since (d/dx) sec x = sec x tan x, we can separate a

sec x tan x factor and convert the remaining (even) power of tangent to secant.

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Example 5

Evaluate

∫

^tan⁶^{x sec}⁴^{x dx.}

Solution:

If we separate one sec²x factor, we can express the

remaining sec²x factor in terms of tangent using the identity sec²x = 1 + tan²x.

We can then evaluate the integral by substituting u = tan x so that du = sec²x dx:

∫

^tan⁶^{x sec}⁴^{x dx =}

∫

^tan⁶^{x sec}²^{x sec}²^{x dx}

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Example 5 – Solution

=

∫

^tan⁶^{x (1 + tan}²^{x) sec}²^{x dx}

=

∫

^u⁶^{(1 + u}²^{)du =}

∫

^(u⁶ ^{+ u}⁸^)du

=

= tan⁷x + tan⁹x + C

cont’d

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Trigonometric Integrals

The preceding examples demonstrate strategies for

evaluating integrals of the form ∫ tan^mx secⁿx dx for two cases, which we summarize here.

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Trigonometric Integrals

For other cases, the guidelines are not as clear-cut. We may need to use identities, integration by parts, and

occasionally a little ingenuity.

We will sometimes need to be able to integrate tan x by using the formula given below:

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Trigonometric Integrals

We will also need the indefinite integral of secant:

We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x + tan x:

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Trigonometric Integrals

If we substitute u = sec x + tan x, then

du = (sec x tan x + sec²x) dx, so the integral becomes

∫

(1/u) du = ln |u| + C.

Thus we have

∫

sec x dx = ln |sec x + tan x | + C

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Example 7

Find

∫

^tan³^{x dx.}

Solution:

Here only tan x occurs, so we use tan²x = sec²x – 1 to rewrite a tan²x factor in terms of sec²x:

∫

^tan³^{x dx =}

∫

^{tan x tan}²^{x dx}

=

∫

^{tan x (sec}²^{x – 1) dx}

=

∫

^{tan x sec}²^{x dx –}

∫

^{tan x dx}

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Example 7 – Solution

= – ln | sec x | + C

In the first integral we mentally substituted u = tan x so that du = sec²x dx.

cont’d

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Trigonometric Integrals

Finally, we can make use of another set of trigonometric identities:

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Example 9

Evaluate

∫

sin 4x cos 5x dx.

Solution:

This integral could be evaluated using integration by parts, but it’s easier to use the identity in Equation 2(a) as follows:

∫

sin 4x cos 5x dx =

∫

[sin(–x) + sin 9x] dx

=

∫

(–sin x + sin 9x) dx

= (cos x – cos 9x) + C