10.4 Integration Of Powers Of Trigonometric Functions
| Calculus Of One Real Variable – By Pheng Kim Ving |
| 10.4 |
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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 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.
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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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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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