Higher_Mathematics_Part_2

3.3. Express the integrant as the sum of partial fractions and determine the resulting integral.

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3.4. Express the integrant as the sum of partial fractions and determine the resulting integral.

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3.5. Express the integrant as the sum of partial fractions and determine the resulting integral.

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Topic 4. Integrals involving powers of trigonometric functions

This section shows how to integrate certain products of powers of the six trigonometric functions, sin x, cos x, tg x (tan x), ctg x (cot x), sec x, cosec x (csc x),

cosec x = sin1 x , any such product can be expressed in the form sinnx cosmx for some integers n and m, positive, zero, or negative. The methods in this section

thus show how to compute ∫ sinn x cosm x dx for certain convenient

combinations of n and m.

The technique described in this section — a particular substitution — reduces the integration of any rational function R of sin x and cos x such as

∫ R(sin x, cos x)dx to the integration of a rational function of t. The latter can be accomplished by partial fractions. The method depends on the fact that sin x and cos x can both be expressed as rational functions of tg 2x .

Consider – π < x < π and let t = tg 2x .

Literature: [1, section 6, ch. 6.5], [2, section 2, ch. 2.1], [4, section 7, § 22], [6, section 8], [7, section 10, § 12], [9, § 32].

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4.1. Integration of sin mx cos nx, sin mx sin nx and cos mx cos nx

In this case are given the formulas:

Consider the integral ∫ sinn x cosm xdx , where n and m are nonnegative integers.

4.2.1.If m = 1 and n ≥ 1 then the integral becomes: ∫ sinn x cos xdx . The substitution t = sin x turns this integral into the easy integral ∫tndt.

4.2.2.If n = 1 and m ≥ 1 then the integral becomes: ∫ cosm xsin xdx .

The substitution t = cos x turns this integral into the easy integral −∫tndt. 4.2.3. If m = 0 and n ≥ 3 and n = 2k + 1 is odd positive integer then the given

integral becomes ∫sinn x dx .

Recall that d(cos x) = –sin x dx. Thus

∫sin2k +1x dx = ∫sin2k x sin xdx = −∫(1− cos 2 x)k d (cos x) . This integral is easier than is given.

4.2.4.Similarly if n = 0 and m ≥ 3 and m = 2k + 1 is odd positive integer:

∫cos2k +1x dx = ∫cos2k x cos xdx = ∫(1− sin 2 x)k d (sin x) .

4.2.5.If m = 0 and n ≥ 2 and n = 2k is even positive integer then the given integral becomes ∫sin2k x dx .

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Similarly replace cos2x by 1+ cos2x . 2

4.2.7. More generally to find ∫ sinn x cosm xdx , where m and n are non-

negative integers and n is odd, pair one sin x with dx to form sin x = − d(cosx) and use the identity sin2x = 1 − cos2x together with the substitution t = cosx. The new integrand will be a polynomial in t. A similar approach works on

∫ sinn x cosm xdx , if m is odd, as is illustrated by example 9.

If both n and m are even in ∫ sinn x cosm xdx the method of example 9 does

facilitate the computation of ∫tgmx secnx dx m and n nonnegative integers, when

m is odd or n is even. When m is odd , form sec x tg x dx , when n is even, form sec2x dx.

4.4. Description of the method

The method depends on the fact that cos x and sin x can both be expressed

This substitution transforms any integral of a rational function of cos x and sin x into an integral of a rational function of t. The resulting rational function can then be integrated by the method of partial fractions.

This method is used to integrate a function such as

∫ dx . a cos x + b sin x + c

Keep in mind that the substitution t = tg 2x is called upon only when easier ways, such as those in the preceding section, don’t work.

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Summary of this section

The following table summarizes the techniques discussed and similar ones for other powers of trigonometric functions.

Table 2.2