121616949-math.173

# 121616949-math.173 - ax b dx = Z 1 a sin u du = 1 a-cos u C...

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8.1 Substitution 159 This is not the only way to do the algebra, and typically there are many paths to the correct answer. Another possibility, for example, is: Since du/dx = 2 x , dx = du/ 2 x , and then the integral becomes Z 2 x cos( x 2 ) dx = Z 2 x cos u du 2 x = Z cos u du. The important thing to remember is that you must eliminate all instances of the original variable x . EXAMPLE 8.1 Evaluate Z ( ax + b ) n dx , assuming that a and b are constants, a = 0, and n is a positive integer. We let u = ax + b so du = a dx or dx = du/a . Then Z ( ax + b ) n dx = Z 1 a u n du = 1 a ( n + 1) u n +1 + C = 1 a ( n + 1) ( ax + b ) n +1 + C. EXAMPLE 8.2 Evaluate Z sin( ax + b ) dx , assuming that a and b are constants and a = 0. Again we let u = ax + b so du = a dx or dx = du/a . Then
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Unformatted text preview: ax + b ) dx = Z 1 a sin u du = 1 a (-cos u ) + C =-1 a cos( ax + b ) + C. EXAMPLE 8.3 Evaluate Z x sin( x 2 ) dx . Let u = x 2 so du = 2 x dx or x dx = du/ 2. Then Z x sin( x 2 ) dx = Z 1 2 sin u du = 1 2 (-cos u ) + C =-1 2 cos( x 2 ) + C. EXAMPLE 8.4 Evaluate Z cos( πt ) sin 2 ( πt ) dt . Let u = sin( πt ) so du = π cos( πt ) dt or du/π = cos( πt ) dt . Then Z cos( πt ) sin 2 ( πt ) dt = Z 1 π 1 u 2 du = Z 1 π u-2 du = 1 π u-1-1 + C =-1 uπ + C =-1 π sin( πt ) + C....
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