Calculus Of One Real Variable – By Pheng Kim Ving
Chapter 10: Techniques Of Integration – Section 10.4: Integration Of Powers Of Trigonometric Functions

10.4
Integration Of Powers Of Trigonometric Functions

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An integer has 2 possibilities for parity: even or odd. So a set of 2 integers say m and n have 2 x 2 = 4 possibilities for
parity, as follows:

m Or n Is Odd

“m or n is odd” means “either m or n is odd or both are odd”.

Example 1.1

Calculate:

Solution

EOS

An integral of the form:

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 utilized.

Both m And n Are Even

Example 1.2

Compute:

Solution

EOS

An integral of the form:

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m Is Even And n Is Either Even Or Odd

Example 2.1

Evaluate:

Solution

EOS

An integral of the form:

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.

Both m And n Are Odd

Example 2.2

Find:

Solution

EOS

An integral of the form:

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

m Is Odd And n Is Even

If we extract sec² x from sec^m 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 sec^m–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 sec^m 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 tan^n–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.

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Example 3.1

Calculate:

Solution

EOS

Integrals of the form:

and the substitution either u = cot x or u = csc x.

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Problems & Solutions

Note

So they differ only by a constant. They're actually equivalent up to different choices of the constant of integration. Recall
that indefinite integrals are general antiderivatives. If both F₁(x) and F₂(x) are particular antiderivatives of f(x), then we
can employ either F₁(x) + C or F₂(x) + C as the general antiderivative of f(x). Any 2 antiderivatives of a function differ
from each other by a constant; see Section 5.7 Part 2. Thus F₁(x) and F₂(x) differ by a constant, and so do F₁(x) + C
and F₂(x) + C.

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

1. Calculate:

Solution

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2. Compute:

Solution

Let v = tan u. Then dv = sec² u du. So:

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3. Evaluate:

Solution

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4. Find:

Solution

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5. Calculate:

Solution

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