week11.pdf - Math 1010 Week 11 Indefinite Integrals Integration of Trig Functions Trigonometric Substitution Integration of Trigonometric Functions We

# week11.pdf - Math 1010 Week 11 Indefinite Integrals...

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Math 1010 Week 11 Indefinite Integrals Integration of Trig. Functions Trigonometric Substitution Integration of Trigonometric Functions We have seen that: Z sin 2 x dx = x 2 - 1 4 sin(2 x ) + C Z cos 2 x dx = x 2 + 1 4 sin(2 x ) + C Example 11.1. Using: Z sec 2 x dx = tan x + C, Z csc 2 x dx = - cot x + C, and the identity 1 + tan 2 x = sec 2 x (which follows from the Pythagorean Theo- rem), we may evaluate: Z tan 2 x dx Z tan 2 x dx = Z (sec 2 x - 1) dx = tan x - x + C, 1
where C represents an arbitrary constant. Z cot 2 x dx Z cot 2 x dx = Z (csc 2 x - 1) dx = - cot x - x + C, where C represents an arbitrary constant. To evaluate an integral of the form: Z sin m x cos n x dx, n, m N , it is useful to make the following substitution: u = ( cos x, if m is odd, sin x, if n is odd, and then apply the Pythagorean Theorem cos 2 x +sin 2 x = 1 to rewrite the original integral as: Z P ( u ) du, where P ( u ) is some polynomial in u . Example 11.2. Evaluate: Z cos 5 x sin 3 x dx Z cos 5 x sin 3 x dx = Z cos 5 x sin 2 x (sin x dx ) Let u = cos x . Then, du = sin x dx . So, Z cos 5 x sin 3 x dx = Z cos 5 x sin 2 x (sin x dx ) = Z u 5 (1 - u 2 ) du = Z ( u 5 - u 7 ) du = 1 6 u 6 - 1 8 u 8 + C = 1 6 cos 6 x - 1 8 cos 8 x + C, where C represents an arbitrary constant.