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EVALUATION OF TRIGONOMETRIC INTEGRALS

Introduction. Many problems lead directly or indirectly to integrals involving powers of
the trigonometric functions. By using the fundamental identities and the double- and half-angle
formulas, one can often transform such integrals into forms in which the standard integration
formulas can be applied. Consider ∫ cos^{3} x dx. None of the formulas applies directly to this
integral, however

∫ cos^{3}x dx = ∫ cos^{2}x cos x dx

= ∫ (1- sin^{2}x) cos x dx

= ∫ cos x dx - ∫ sin^{2} x cos x dx

= sin x - ⅓ sin^{3} x + c

We shall now give techniques for evaluating certain types of trigonometric integrals which arise often in applications and for which simple rules for integration can be given — namely the following types:

∫ sin^{m} x cos^{n} x dx

∫ sec^{n} x dx, ∫ csc^{n} x dx

∫ tan^{n} x dx, ∫ cot^{n} x dx

∫ tan^{m} x sec^{n} x dx, ∫ cot^{m} x csc^{n} x dx

∫ sin mx cos nx dx, ∫ sin mx sin nx dx, ∫ cos mx cos nx dx

I* *Evaluating integrals of type ∫ sin^{m }x cos^{n} x dx .

Case 1. One of the exponents is a positive odd integer. Suppose for example that m is odd. One can then take out sin x dx as du, leaving an even exponent for sin x. Then, using the relation

sin^{2} x = 1 - cos^{2} x

he can obtain a series of terms of the form

∫ cos^{q} x sin x dx

which can be integrated by the formula for ∫ u^{n}du.

Example. Find ∫ sin^{3}x cos^{4} x dx.

Solution. Since the exponent of sin x is a positive odd integer, we proceed as follows:

∫ sin^{3}x cos^{4} x dx = ∫ cos^{4} x sin^{2} x sin x dx

= ∫ cos^{4} x(1 - cos^{2} x) sin x dx

= ∫ cos^{4} x sin x dx - ∫ cos^{6} x sin x dx

Note. Note that this procedure will always apply if one of the exponents is a positive odd integer no matter what the other exponent may be i.e. the other might be any positive or negative integer or fraction, or zero.

Case 2. Both exponents are positive even integers. The integration in this case can be accomplished by changing over to multiple angles. To change over to multiple angles we utilize the following trigonometric formulas:

sin^{2} x = ½ (1 - cos 2x)

cos^{2} x = ½ (1 + cos 2x)

sin x cos x = ½ sin 2x

Example 1. Find ∫ cos^{2} x dx.

Solution.

∫ cos^{2} x dx = ∫ ½ (1 + cos 2x) dx

= ½ ∫dx + ½ ∫cos 2x dx

= ½ x + ¼ sin 2x + c

Example 2. Find ∫ sin^{2} x cos^{2} x dx.

Solution.

∫sin^{2} x cos^{2} x dx = ¼ ∫ sin^{2} 2x dx

= ¼ ∫ ½ (1 - cos 4x)dx

II Evaluating integrals of type ∫ sec^{n} x dx or ∫ csc^{n} x dx .

Case 1. Exponent n is a positive even integer. To evaluate ∫ sec^{n} x dx take out sec^{2} x
dx as du and transform the remainder into a polynomial in tan x. To evaluate ∫ csc^{n} x dx take
out csc^{2} x dx as du and transform the remainder into a polynomial in cot x.

Example. Find ∫ sec^{6} x dx.

Solution.

∫ sec^{6} x dx = ∫ sec^{4} x sec^{2} x dx

= ∫(1 + tan^{2} x)^{2} sec^{2} x dx

= ∫(1 + 2 tan^{2} x + tan^{4} x)sec^{2} x dx

= ∫ sec^{2} x dx + 2 ∫ tan^{2} x sec^{2} x dx + ∫ tan^{4} x sec^{2} x dx

Case 2. Exponent n is a positive odd integer. Use integration by parts technique to integrate.

III Evaluating integrals of type ∫ tan^{n} x dx or ∫ cot^{n} x dx where n is any
positive integer.

Reduce the integrals to forms that are easily integrated by use of the trigonometric identities

tan^{2} x = sec^{2} x - 1

cot^{2} x = csc^{2} x - 1

Example. Find ∫ tan^{4} x dx.

Solution.

∫ tan^{4} x dx = ∫ tan^{2} x (sec^{2} x - 1)dx

= ∫ tan^{2} x sec^{2} x dx - ∫ tan^{2} x dx

= ⅓ tan^{3} x - ∫(sec^{2} x - 1)dx

= ⅓ tan^{3} x - tan x + x + c

IV* *Evaluating integrals of type ∫ tan^{m }x sec^{n} x dx or ∫ cot^{m }x csc^{n} x dx . We
will give the treatment of ∫ tan^{m} x sec^{n} x dx. The treatment for ∫ cot^{m} x csc^{n} x dx is analogous.

Case 1. Exponent of sec x is a positive even integer. Take out sec^{2} x dx as du and
transform the remainder into a polynomial in tan x.

Example. Find ∫ tan^{3} x sec^{4} x dx.

Solution. Since the exponent of sec x is a positive even integer, we may proceed as follows:

∫ tan^{3} x sec^{4} x dx = ∫ tan^{3} x sec^{2} x sec^{2} x dx

= ∫ tan^{3} x(1 + tan^{2} x) sec^{2} x dx

= ∫ tan^{3} x sec^{2} x dx + ∫ tan^{5} x sec^{2} x dx

Case 2. Exponent of tan x is a positive odd integer. Take out sec x tan x dx as du and transform the remainder into a polynomial in sec x.

Example. Find ∫ sec^{3} x tan^{3} x dx.

Solution. Since the exponent of tan x is a positive odd integer, we proceed as follows:

∫ sec^{3} x tan^{3} x dx = ∫ sec^{2} x tan^{2} x sec x tan x dx

= ∫ sec^{2} x (sec^{2} x - 1) sec x tan x dx

= ∫ sec^{4} x sec x tan x dx - ∫ sec^{2} x sec x tan x dx

It is possible that a given integral could be handled by either of these procedures. Thus, ∫ sec^{4} x
tan^{3} x dx could be found by either method, since the exponent of sec x is even and that of tan x is
odd. On the other hand, neither of the procedures could be used on ∫ sec^{3} x tan^{2} x dx.

V* *Evaluating integrals of type ∫ sin mx cos nx dx , ∫ sin mx sin nx dx and ∫
cos mx cos nx dx . These integrals are given by the following formulas for m
n.

References.

Middlemiss. Differential and Integral Calculus. Chap. 16.

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