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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 ∫ cos3 x dx. None of the formulas applies directly to this integral, however

∫ cos3x dx = ∫ cos2x cos x dx

= ∫ (1- sin2x) cos x dx

= ∫ cos x dx - ∫ sin2 x cos x dx

= sin x - ⅓ sin3 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:

∫ sinm x cosn x dx

∫ secn x dx, ∫ cscn x dx

∫ tann x dx, ∫ cotn x dx

∫ tanm x secn x dx, ∫ cotm x cscn x dx

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

I Evaluating integrals of type sinm x cosn 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

sin2 x = 1 - cos2 x

he can obtain a series of terms of the form

∫ cosq x sin x dx

which can be integrated by the formula for ∫ undu.

Example. Find ∫ sin3x cos4 x dx.

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

∫ sin3x cos4 x dx = ∫ cos4 x sin2 x sin x dx

= ∫ cos4 x(1 - cos2 x) sin x dx

= ∫ cos4 x sin x dx - ∫ cos6 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:

sin2 x = ½ (1 - cos 2x)

cos2 x = ½ (1 + cos 2x)

sin x cos x = ½ sin 2x

Example 1. Find ∫ cos2 x dx.

Solution.

∫ cos2 x dx = ∫ ½ (1 + cos 2x) dx

= ½ ∫dx + ½ ∫cos 2x dx

= ½ x + ¼ sin 2x + c

Example 2. Find ∫ sin2 x cos2 x dx.

Solution.

∫sin2 x cos2 x dx = ¼ ∫ sin2 2x dx

= ¼ ∫ ½ (1 - cos 4x)dx

II Evaluating integrals of type secn x dx or cscn x dx .

Case 1. Exponent n is a positive even integer. To evaluate ∫ secn x dx take out sec2 x dx as du and transform the remainder into a polynomial in tan x. To evaluate ∫ cscn x dx take out csc2 x dx as du and transform the remainder into a polynomial in cot x.

Example. Find ∫ sec6 x dx.

Solution.

∫ sec6 x dx = ∫ sec4 x sec2 x dx

= ∫(1 + tan2 x)2 sec2 x dx

= ∫(1 + 2 tan2 x + tan4 x)sec2 x dx

= ∫ sec2 x dx + 2 ∫ tan2 x sec2 x dx + ∫ tan4 x sec2 x dx

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

III Evaluating integrals of type tann x dx or cotn x dx where n is any positive integer.

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

tan2 x = sec2 x - 1

cot2 x = csc2 x - 1

Example. Find ∫ tan4 x dx.

Solution.

∫ tan4 x dx = ∫ tan2 x (sec2 x - 1)dx

= ∫ tan2 x sec2 x dx - ∫ tan2 x dx

= ⅓ tan3 x - ∫(sec2 x - 1)dx

= ⅓ tan3 x - tan x + x + c

IV Evaluating integrals of type tanm x secn x dx or cotm x cscn x dx . We will give the treatment of ∫ tanm x secn x dx. The treatment for ∫ cotm x cscn x dx is analogous.

Case 1. Exponent of sec x is a positive even integer. Take out sec2 x dx as du and transform the remainder into a polynomial in tan x.

Example. Find ∫ tan3 x sec4 x dx.

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

∫ tan3 x sec4 x dx = ∫ tan3 x sec2 x sec2 x dx

= ∫ tan3 x(1 + tan2 x) sec2 x dx

= ∫ tan3 x sec2 x dx + ∫ tan5 x sec2 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 ∫ sec3 x tan3 x dx.

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

∫ sec3 x tan3 x dx = ∫ sec2 x tan2 x sec x tan x dx

= ∫ sec2 x (sec2 x - 1) sec x tan x dx

= ∫ sec4 x sec x tan x dx - ∫ sec2 x sec x tan x dx

It is possible that a given integral could be handled by either of these procedures. Thus, ∫ sec4 x tan3 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 ∫ sec3 x tan2 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.