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CALCULUS 1 PROBLEMS & SOLUTIONS

6.5.1.3

Integration Of Powers Of Trigonometric Functions

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Review

1. The Integral ò sinm x cosn x dx Where Either m Or n Is Odd

An integral of the form:

ò sinm x cosn x dx,

where either m or n is an odd positive integer, can be found easily by the method of substitution and

the use of the trigonometric identity sin2 x + cos2 x = 1, as follows.

Suppose m is odd. Then m = 2k + 1, where k is a non-negatve integer. Hence:

ò sinm x cosn x dx = ò sin2k+1 x cosn x dx = ò (sin2 x)k cosn x sin x dx = ò (1 – cos2 x)k cosn x sin x dx.

Let u = cos x. Then du = – sin x dx. Thus:

ò sinm x cosn x dx = – ò (1 – u2)k un du.

We’ve got the integral of a polynomial in u, which can readily be found. Don’t forget to return to the

original variable x. Similarly, if n is odd, then the substitution u = sin x can be employed.

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2. The Integral ò sinm x cosn x dx Where Both m And n Are Even

An integral of the form:

ò sinm x cosn x dx,

where both m and n are even non-negative integers, can be computed by using the trigonometric

identities sin2 x = (1 – cos 2x)/2 and cos2 x = (1 + cos 2x)/2, as follows:

We have m = 2k and n = 2l, where k and l are non-negative integers. Then:

ò sinm x cosn x dx = ò (sin2 x)k (cos2 x)l dx = ò ((1 – cos 2x)/2)k ((1 + cos 2x)/2)l dx.

Multiply out ((1 – cos 2x)/2)k and ((1 + cos 2x)/2)l. We use the two identities sin2 x = (1 – cos 2x)/2

and cos2 x = (1 + cos 2x)/2 repeatedly if necessary until we get just odd powers of cos x and

constants, for which we can use the formula ò cos ax dx = (1/a) sin ax + C and the method

presented in part 1.

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3. The Integral ò secm x tann x dx Where m Is Even

An integral of the form:

ò secm x tann x dx,

where m is an even positive integer, can be done by the method of substitution and the use of the

identity 1 + tan2 x = sec2 x, as follows.

We have m = 2k, where k is a positive integer. Then:

ò secm x tann x dx = ò (sec2 x)k–1 tann x sec2 x dx = ò (1 + tan2 x)k–1 tann x sec2 x dx.

Let u = tan x. So du = sec2 x dx. It follows that:

ò secm x tann x dx = ò (1 + u2)k–1 un du.

We’ve obtained the integral of a polynomial in u, which can readily be done. Don’t forget to return to

the original variable x.

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4. The Integral ò secm x tann x dx Where Both m And n Are Odd

An integral of the form:

ò secm x tann x dx,

where both m and n are odd positive integers, can be computed by the method of substitution and the

utilization of the identity 1 + tan2 x = sec2 x, as follows.

We have m = 2k + 1 and n = 2l + 1, where k and l are non-negative integers. Then:

ò secm x tann x dx = ò sec2k x (tan2 x)l sec x tan x dx = ò sec2k x (sec2 x – 1)l sec x tan x dx.

Let u = sec x. So du = sec x tan x dx. Thus:

ò secm x tann x dx = ò u2k (u2 – 1)l du.

We’ve got the integral of a polynomial in u, which can handily be computed. Then we return to the

original variable x.

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5. The Integral ò secm x tann x dx Where m Is Odd And n Is Even

If we extract sec2 x from secm x to form sec2 x dx because the derivative of tan x is sec2 x, then we

would have to make the substitution u = tan x, so secm–2 x would have to be changed to an expression

involving only integer powers of tan x and constants using the identity 1 + tan2 x = sec2 x, which is

impossible because m – 2 is odd.

If we extract sec x tan x from secm x tann x to form sec x tan x dx because the derivative of sec x is

sec x tan x, then we would have to make the substitution u = sec x, so tann–1 x would have to be

changed to an expression involving only integer powers of sec x and constants using the identity

1 + tan2 x = sec2 x, which is impossible because n – 1 is odd.

An integral of the form ò secm x tann x dx, where m is odd and n is even, can be handled by the

method of integration by parts. This technique will be discussed in a later section.

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6. The Integral ò cscm x cotn x dx

Integrals of the form ò cscm x cotn x dx can be done in a similar fashion to integrals of the form

ò secm x tann x dx, using the following features: (d/dx) cot x = – csc2 x, (d/dx) csc x =

– csc x cot x, identity 1 + cot2 x = csc2 x, and substitution either u = cot x or u = csc x.

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Problems & Solutions

Note

The following problems ask you to find integrals. For indefinite integrals, it may occur that your answers

appear different from the ones given in the solutions while they’re still correct. For example, let’s find

ò sin x cos x dx.

Using the substitution u = sin x, which yields du = cos x dx, we have ò sin x cos x dx = ò u du =

(1/2) u2 + C = (1/2) sin2 x + C. Now using the substitution u = cos x, which yields du = – sin x dx,

we have ò sin x cos x dx = – ò u du = – (1/2) u2 + C = – (1/2) cos2 x + C. Differentiating both

answers will confirm that they’re both correct.

Let’s find out what’s the difference between the two answers: (1/2) sin2 x – (– (1/2) cos2 x) =

(1/2) (sin2 x + cos2 x) = (1/2) (1) = 1/2. Thus, they differ only by a constant, whose derivative is 0.

The answers are actually equivalent up to different choices of the constant of integration.

If your answer looks different from the one provided, then just differentiate it to see if it’s also a

correct one.

1. Find the following integrals.

Solution

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2. Find the following integrals.

a. ò sec2 u tan4 u du.

b. ò sec5 y tan y dy.

Solution

a. Let v = tan u. Then dv = sec2 u du. Thus:

ò sec2 u tan4 u du = ò v4 dv = (1/5) v5 + C = (1/5) tan5 u + C.

b. ò sec5 y tan y dy = ò sec4 y sec y tan y dy.

Let u = sec y. Then du = sec y tan y dy. Therefore:

ò sec5 y tan y dy = ò u4 du = (1/5) u5 + C = (1/5) sec5 y + C.

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3. Find the following integrals.

a. ò csc4 x cot x dx.

b. ò csc5 r cot5 r dr.

Solution

a. ò csc4 x cot x dx = ò csc2 x cot x csc2 x dx

= ò (1 + cot2 x) cot x csc2 x dx

= ò (cot x + cot3 x) csc2 x dx.

Let u = cot x. Then du = – csc2 x dx. Thus:

ò csc4 x cot x dx = – ò (u + u3) du

= – (1/2) u2 – (1/4) u4 + C

= – (1/2) cot2 x – (1/4) cot4 x + C.

b. ò csc5 r cot5 r dr = ò csc4 r cot4 r csc r cot r dr

= ò csc4 r (cot2 r)2 csc r cot r dr

= ò csc4 r (csc2 r – 1)2 csc r cot r dr

= ò (csc8 r – 2 csc6 r + csc4 r) csc r cot r dr.

Let u = csc r. Then du = – csc r cot r dr. Thus:

ò csc5 r cot5 r dr = – ò (u8 – 2u6 + u4) du

= – (1/9) u9 + (2/7) u7 – (1/5) u5 + C

= – (1/9) csc9 r + (2/7) csc7 r – (1/5) csc5 r + C.

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Solution

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# 6.5.1.3 Integration Of Powers Of Trigonometric Functions

Trigonometric Integrals – Part 6 of 6

Trigonometric Integrals – Part 6 of 6