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6.3 integration by substitution

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This document discusses integration by substitution. It provides an example of recognizing a composite function and rewriting the integral in terms of the inside and outside functions. Specifically, it shows rewriting the integral of (x2 +1)2x dx as the integral of the outside function (x2 + 1) with the inside function (x) plugged in, plus a constant. It then provides additional practice problems applying the technique of substitution to rewrite integrals in terms of u-substitutions.

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Integration by Substitution
Recognizing the Outside-Inside Pattern (x 2 +1) 2x dx 2 From doing derivatives we need to recognize the integrand above is a composite function from the derivative of the outside times the derivative of the inside (chain rule). 3 1 2 = ( x + 1) + C 3 + C since this is an indefinite integral
Think of this function as 2 functions: f(x) and g(x) f ( x) = x g ( x) = x + 1 2 2 As a composite function then: ( ) f ( g ( x) ) = x + 1 outside 2 2 inside Now look at the original integral: (x 2 +1) 2x dx f(g(x)) 2 g(x)
Read this as the antiderivative of the outside function with the inside function plugged inplus C
6.3 integration by substitution
Lets Practice !!! 3x 2 sin x 3 dx = 3x 2 sin x 3 dx = du Let u = x3 du = 3x2 sin u sin u du = cosu + C = cos x + C 3
More Practice !!! x 1/4 3 x + 2 dx = 4 Let u = x4 + 2 (x 4 x 4 + 2 x 3 dx = 1 4 u du 1 4 du = 4 x3 4 u du = 3 2 3 2 1u 1 2u u du = 4 3 + C = 4 3 + C = 2 + 2) 6 1 2 3 2 +C = (x 4 + 2) 6 3 +C
Here are some problems for you to work on!!! dx 1. = 2 1+ 3x 錚1 2 錚 2. 錚 sin x 歎dx = 錚x 錚 3. sin 2 x cos x dx = e x 4. dx = x 43 5 5. t 3 5t dt =
Less Apparent Substitution 1. x 2 x 1 dx = x 2 x 1 dx = (u + 1) Let u = x 1 du = dx x=u+1 x2 = (u + !)2 7 2 ( u +1) ( 5 2 u 2 du u du = ( u +1) 2 1 2 u du = 3 1錚 錚 5 2 錚 u 2 + 2u 2 + u 2 歎du = u + 2u +1 u du = 錚 錚 ) 3 2 1 2 7 2 5 2 3 2 2 ( x 1) 4 ( x 1) 2 ( x 1) 2u 2 2u 2u + + +C = + + +C 7 5 3 7 5 3
Less Apparent Substitution 1. x 2 x 1 dx = x 2 x 1 dx = (u + 1) Let u = x 1 du = dx x=u+1 x2 = (u + !)2 7 2 ( u +1) ( 5 2 u 2 du u du = ( u +1) 2 1 2 u du = 3 1錚 錚 5 2 錚 u 2 + 2u 2 + u 2 歎du = u + 2u +1 u du = 錚 錚 ) 3 2 1 2 7 2 5 2 3 2 2 ( x 1) 4 ( x 1) 2 ( x 1) 2u 2 2u 2u + + +C = + + +C 7 5 3 7 5 3

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6.3 integration by substitution

  • 2. Recognizing the Outside-Inside Pattern (x 2 +1) 2x dx 2 From doing derivatives we need to recognize the integrand above is a composite function from the derivative of the outside times the derivative of the inside (chain rule). 3 1 2 = ( x + 1) + C 3 + C since this is an indefinite integral
  • 3. Think of this function as 2 functions: f(x) and g(x) f ( x) = x g ( x) = x + 1 2 2 As a composite function then: ( ) f ( g ( x) ) = x + 1 outside 2 2 inside Now look at the original integral: (x 2 +1) 2x dx f(g(x)) 2 g(x)
  • 4. Read this as the antiderivative of the outside function with the inside function plugged inplus C
  • 6. Lets Practice !!! 3x 2 sin x 3 dx = 3x 2 sin x 3 dx = du Let u = x3 du = 3x2 sin u sin u du = cosu + C = cos x + C 3
  • 7. More Practice !!! x 1/4 3 x + 2 dx = 4 Let u = x4 + 2 (x 4 x 4 + 2 x 3 dx = 1 4 u du 1 4 du = 4 x3 4 u du = 3 2 3 2 1u 1 2u u du = 4 3 + C = 4 3 + C = 2 + 2) 6 1 2 3 2 +C = (x 4 + 2) 6 3 +C
  • 8. Here are some problems for you to work on!!! dx 1. = 2 1+ 3x 錚1 2 錚 2. 錚 sin x 歎dx = 錚x 錚 3. sin 2 x cos x dx = e x 4. dx = x 43 5 5. t 3 5t dt =
  • 9. Less Apparent Substitution 1. x 2 x 1 dx = x 2 x 1 dx = (u + 1) Let u = x 1 du = dx x=u+1 x2 = (u + !)2 7 2 ( u +1) ( 5 2 u 2 du u du = ( u +1) 2 1 2 u du = 3 1錚 錚 5 2 錚 u 2 + 2u 2 + u 2 歎du = u + 2u +1 u du = 錚 錚 ) 3 2 1 2 7 2 5 2 3 2 2 ( x 1) 4 ( x 1) 2 ( x 1) 2u 2 2u 2u + + +C = + + +C 7 5 3 7 5 3
  • 10. Less Apparent Substitution 1. x 2 x 1 dx = x 2 x 1 dx = (u + 1) Let u = x 1 du = dx x=u+1 x2 = (u + !)2 7 2 ( u +1) ( 5 2 u 2 du u du = ( u +1) 2 1 2 u du = 3 1錚 錚 5 2 錚 u 2 + 2u 2 + u 2 歎du = u + 2u +1 u du = 錚 錚 ) 3 2 1 2 7 2 5 2 3 2 2 ( x 1) 4 ( x 1) 2 ( x 1) 2u 2 2u 2u + + +C = + + +C 7 5 3 7 5 3