121616949-math.181 - x dx =-x cos x sin x C EXAMPLE 8.13...

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8.4 Integration by Parts 167 but that Z f ( x ) g ( x ) dx is easier. This technique for turning one integral into another is called integration by parts , and is usually written in more compact form. If we let u = f ( x ) and v = g ( x ) then du = f ( x ) dx and dv = g ( x ) dx and Z u dv = uv - Z v du. To use this technique we need to identify likely candidates for u = f ( x ) and dv = g ( x ) dx . EXAMPLE 8.11 Evaluate Z x ln x dx . Let u = ln x so du = 1 /x dx . Then we must let dv = x dx so v = x 2 / 2 and Z x ln x dx = x 2 ln x 2 - Z x 2 2 1 x dx = x 2 ln x 2 - Z x 2 dx = x 2 ln x 2 - x 2 4 + C. EXAMPLE 8.12 Evaluate Z x sin x dx . Let u = x so du = dx . Then we must let dv = sin x dx so v = - cos x and Z x sin x dx = - x cos x - Z - cos x dx = - x cos x +
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Unformatted text preview: x dx =-x cos x + sin x + C. EXAMPLE 8.13 Evaluate Z sec 3 x dx . Of course we already know the answer to this, but we needed to be clever to discover it. Here we’ll use the new technique to discover the antiderivative. Let u = sec x and dv = sec 2 x dx . Then du = sec x tan x and v = tan x and Z sec 3 x dx = sec x tan x-Z tan 2 x sec x dx = sec x tan x-Z (sec 2 x-1) sec x dx = sec x tan x-Z sec 3 x dx + Z sec x dx....
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