Methods of Integration Professor D. Olles January 8, 04 Substitution The derivative of a composition of functions can be found using the chain rule form d dx [f (g(x))] f (g(x)) g (x) Rewriting the derivative form in the differential form give us d [f (g(x))] f (g(x)) g (x)dx Integrating both sides of the equation results in d [f (g(x))] f (g(x)) g (x)dx f (g(x)) + C f (g(x)) g (x)dx Let Then u g(x) du dx g (x) du g (x)dx The above equation can be written as f (u)du f(u) + C Or, if the integrand is some function y f(g(x)), f(u)du F (u) + C Note that in the case of the definite integral, it is necessary to convert our limits of integration to match up with the new independent variable. That is: if x a then u g(a) if x b then u g(b)
and the integral formula becomes: b a f(g(x))g (x)dx g(b) g(a) f(u)du F (u) g(b) g(a) F (g(b)) F (g(a)) Example. substitution. Evaluate the following indefinite integral using the method of e x dx u x du dx du dx e x dx e u du e u du eu + C ex + C Example. Evaluate the following indefinite integral using the method of substitution. x + x + x dx u x + x du (x + )dx (x + )dx du (x + )dx x + x + x dx u du u du ln u + C ln x + x + C
Example. substitution. Example.4 substitution. Evaluate the following indefinite integral using the method of e arctan x + x dx u arctan x du + x dx e arctan x + x dx e u du e u + C e arctan x + C Evaluate the following definite integral using the method of e sin (ln x) dx x u ln x e du x dx x u ln 0 x e u ln e sin (ln x) dx x 0 sin udu cos u 0 cos () cos (0) cos () + cos ()
Integration by Parts The derivative of the product of two function can be found using the form d dx [f(x)g(x)] f (x)g(x) + f(x)g (x) If we rewrite the derivative form in the differential form, we have d [f(x)g(x)] [f (x)g(x) + f(x)g (x)] dx Integrating both sides allows us to solve this differential equation. d [f(x)g(x)] [f (x)g(x) + f(x)g (x)]dx f(x)g(x) f (x)g(x)dx + f(x)g (x)dx Solving for the second term on the right hand side yields f(x)g (x)dx f(x)g(x) f (x)g(x)dx Let u f(x) and dv g (x)dx Then du f (x)dx and v g(x) Our equation above can be written in simpler form. That is udv uv vdu f(x)g (x)dx f(x)g(x) f (x)g(x)dx In the case of the definite integral, we have b a b f(x)g (x)dx f(x)g(x) b a f (x)g(x)dx Note that since we are not actually performing a change of variables, it is not necessary to change our limits of integration. a 4
Example. Evaluate the following integral using Integration by Parts. x e x dx u x du xdx x e x dx ( x ) ( ex dv e x dx v ex ) ( ex x e x xe x dx ) xdx u x dv e x dx du dx v ex [ ] x e x xex ex dx x e x xex + 4 ex + C Example. Evaluate the following integral using Integration by Parts. e z cos zdz w z dw dz dw dz e z cos zdz e w cos wdw u e w du e w dw dv cos wdw v sin w e z cos zdz e w cos wdw [ e w sin w ] e w sin wdw u e w du e w dw dv sin wdw v cos w 5
[ ( e w sin w e w cos w )] e w ( cos w)dw e w cos wdw e w cos wdw ew sin w + ew cos w e w cos wdw e w cos wdw ew sin w + ew cos w + C e w cos wdw 6 ew sin w + 6 ew cos w + C e z cos zdz 6 e z sin z + 6 e z cos z + C Example. Evaluate the following integral using Integration by Parts. x arctan xdx u arctan x dv x dx du +x dx v x x arctan xdx x arctan x x + x dx u + x du xdx du xdx x du x dx u du x dx x arctan xdx x arctan x u du u x arctan x ( ) du 6 u x arctan x [u ln u ] + C 6 x arctan x [ + x ln ( + x ) ] + C 6 6
x arctan x 6 + 6 x 6 ln ( + x ) + C x arctan x + 6 x 6 ln ( + x ) + C Example.4 Evaluate the following integral using Integration by Parts. ln x x dx u ln x du x dx dv x dx v x ln x x dx ln x x + x dx ln x x x + C Example.5 Evaluate the following integral using Integration by Parts. u ln x du x dx e x ln xdx dv x dx v x x e ln x e x dx e 0 x e 9 e e 9 + 9 e + 9 7
Trigonometric Substitution. Utilizing the Pythagorean Identity sin θ cos θ Consider the Pythagorean Identity sin θ + cos θ and its alternate form sin θ cos θ If we choose to multiply both sides of the equation by a perfect square constant a we have a ( sin θ) a cos θ a a sin θ a cos θ Since the RHS is a product of perfect squares, taking the square root of both sides would reduce the power simply. a a sin θ a cos θ a a sin θ a cos θ If we let x a sin θ, then x a sin θ and a x a cos θ So, the substitution we should make to change our variables is x a sin θ dx a cos θdθ The resulting integrand will be a function of θ yielding an antiderivative as a function of θ. It is then necessary to perform a backsubstitution in order to write our answer in terms of the independent variable x. We do this by returning to the claim that x a sin θ Then sin θ x a OP P HY P and all other 5 trigonometric evaluations can be obtained from the resulting triangle below. 8
Example. Evaluate the following integral. x dx 4 x x sin θ dx cos θdθ sin θ cos θdθ 4 ( sin θ) sin θ cos θdθ 4 4 sin θ sin θ cos θdθ 4( sin θ) sin θ cos θdθ 4 cos θ sin θ cos θdθ cos θ sin θdθ cos θ + C x sin θ sin θ x cos θ 4 x 4 x + C 4 x + C 9
. Utilizing the Pythagorean Identity sec θ tan θ Consider the Pythagorean Identity sin θ + cos θ and its alternate form sec θ tan θ If we choose to multiply both sides of the equation by a perfect square constant a we have a (sec θ ) a tan θ a sec θ a a tan θ Since the RHS is a product of perfect squares, taking the square root of both sides would reduce the power simply. a sec θ a a tan θ a sec θ a a tan θ If we let x a sec θ, then x a sec θ and x a a tan θ So, the substitution we should make to change our variables is x a sec θ dx a sec θ tan θdθ The resulting integrand will be a function of θ yielding an antiderivative as a function of θ. It is then necessary to perform a backsubstitution in order to write our answer in terms of the independent variable x. We do this by returning to the claim that x a sec θ Then sec θ x a HY P ADJ and all other 5 trigonometric evaluations can be obtained from the resulting triangle below. 0
Example. Evaluate the following integral. 4x 9 dx x (x) 9 dx x u x u x du dx du dx u 9 ( u) du u 9 ( du 4u) u 9 du u u sec θ du sec θ tan θdθ ( sec θ) 9 ( sec θ) sec θ tan θdθ 9 sec θ 9 9 sec sec θ tan θdθ θ 9(sec θ ) tan θdθ sec θ 9 tan θ tan θdθ sec θ tan θ tan θdθ sec θ tan θ sec θ dθ sin θ cos cos θdθ θ
sin θ cos θ dθ cos θ dθ ( cos θ ) cos θ cos θ dθ (sec θ cos θ)dθ [ln sec θ + tan θ sin θ] + C u sec θ sec θ u u 9 tan θ u 9 sin θ u [ u ln + u 9 ] u 9 + C u [ x ln + 4x 9 ] 4x 9 + C x x ln + 4x 9 4x 9 + C x
. Utilizing the Pythagorean Identity + tan θ sec θ Consider the Pythagorean Identity sin θ + cos θ and its alternate form + tan θ sec θ If we choose to multiply both sides of the equation by a perfect square constant a we have a ( + tan θ) a sec θ a + a tan θ a sec θ Since the RHS is a product of perfect squares, taking the square root of both sides would reduce the power simply. a + a tan θ a sec θ a + a tan θ a sec θ If we let x a tan θ, then x a tan θ and a + x a sec θ So, the substitution we should make to change our variables is x a tan θ dx a sec θdθ The resulting integrand will be a function of θ yielding an antiderivative as a function of θ. It is then necessary to perform a backsubstitution in order to write our answer in terms of the independent variable x. We do this by returning to the claim that x a tan θ Then tan θ x a OP P ADJ and all other 5 trigonometric evaluations can be obtained from the resulting triangle below.
Example. Evaluate the following integral. x dx 9 + x x tan θ dx sec θdθ ( tan θ) 9 + ( tan θ) sec θdθ 9 tan θ 9 + 9 tan θ sec θdθ 9 tan θ 9( + tan θ) sec θdθ 9 tan θ 9 sec θ sec θdθ 9 tan θ sec θ sec θdθ 9 tan θ sec θdθ 9 tan θ sec θ tan θdθ u tan θ dv sec θ tan θdθ du sec θdθ v sec θ [ ] 9 sec θ tan θ sec θdθ [ ( 9 sec θ tan θ sec θ tan θ + )] ln sec θ + tan θ + C 9 sec θ tan θ 9 ln sec θ + tan θ + C x tan θ tan θ x 4
9 + x sec θ ( ) 9 9 + x (x ) 9 9 + x ln + x + C x 9 + x 9 9 + ln x + x + C 5
4 Partial Fraction Decomposition & Polynomial Long Division When integration rational functions of the form R(x) N(x) D(x), where N and D are polynomials of degree m and n respectively, it may be necessary to perform some algebraic manipulation to rewrite the function in a simpler form, or one whose antiderivative is easier to recognize. That is, of course, if any other method has been exhausted. Consider rational functions whose antiderivatives are not recognizable immediately and cannot be integrated by means of substitution. Let R(x) N(x) D(x) a mx m + a m x m + + a x + a x + a 0 b n x n + b n x n + + b x + b x + b 0 4. m n Long Division of Polynomials Let us review the alternate form division algorithm n(x) r(x) q(x) + d(x) d(x) where n(x) is the numerator of the rational expression, d(x) is the denominator, q(x) is the resulting quotient and r(x) is any remainder as a function of x whose degree is less than the degree of the denominator. We will use an example to review this concept. Example 4. Evaluate the following integral. x x + dx x + ) x x x x x x + x ( x + dx x + ) dx x + (x )dx + x + dx u x + du dx 6
x x + u du x x + ln u + C x x + ln x + + C 4. m < n Partial Fraction Decomposition 4.. Unique Linear Factors If the denominator is the product of unique linear factors, we may rewrite the rational expression as N(x) (ax + b)(cx + d) (ex + f) A ax + b + B cx + d + + C ex + f where A, B,..., C are undetermined constants. We may then algebraically solve for these constants. Example 4. x + x + 5x + 6 dx x + x + 5x + 6 x + (x + )(x + ) A x + + B x + x + A(x + ) + B(x + ) (x + )(x + ) (x + )(x + ) x + Ax + A + Bx + B x + (A + B)x + (A + B) A + B A + B A B A + ( A) A + A A A B B x + x + 5x + 6 x + x + ( x + x + x + x + 5x + 6 dx ln x + ln x + + C ) dx 7
4.. Repeated Linear Factors If the denominator is the product of repeated linear factors, we may rewrite the rational expression as N(x) (ax + b) n A ax + b + B (ax + b) + + C (ax + b) n where A, B,..., C are undetermined constants. We may then algebraically solve for these constants. Example 4. x x + 4x + 4 dx x x + 4x + 4 x (x + ) A x + + B (x + ) x A(x + ) + B (x + ) (x + ) x Ax + A + B A A + B () + B 5 B x x + 4x + 4 x + 5 x x + 4x + 4 dx (x + ) ( x + 5 (x + ) ln x + + 5 x + + C ) dx 4.. Unique Quadratic Factors If the denominator is the product of unique quadratic factors, we may rewrite the rational expression as N(x) (ax + bx + c)(dx + ex + f) (gx + hx + i) Ax + B ax + bx + c + Cx + D dx + + Ex + F + ex + f gx + hx + i Example 4.4 x x + 5 (x + )(x + 4) dx 8
x x + 5 (x + )(x + 4) Ax + B x + + Cx + D x + 4 x x + 5 (x + )(x + 4) (Ax + B)(x + 4) + (Cx + D)(x + ) (x + )(x + 4) x x + 5 Ax + Bx + 4Ax + 4B + Cx + Dx + Cx + D x x + 5 (A + C)x + (B + D)x + (4A + C)x + (4B + D) A + C 0 B + D 4A + C 4B + D 5 A C B D 4C + C 4( D) + D 5 C 4 D 5 A B 4 C D x x + 5 (x + )(x + 4) x + 4 x + + x x + 4 x ( x + 5 (x + )(x + 4) dx x x + + 4/ x + + x x + 4 / ) x dx + 4 x x + dx + 4 x + dx + x x + 4 dx x + 4 dx u x + w x + 4 x tan θ u xdx dw xdx dx sec θdθ du dx dw dx u du + 4 x + dx + w dw ( tan θ) + 4 sec θdθ ln u + 4 arctan x + ln w dθ 6 ln (x + ) + 4 arctan x + ln (x + 4) 6 θ + C ln (x + ) + 4 arctan x + ln (x + 4) 6 arctan ( x ) + C 9
4..4 Repeated Quadratic Factors If the denominator is the product of repeated quadratic factors, we may rewrite the rational expression as N(x) (ax + bx + c) n Example 4.5 Ax + B ax + bx + c + Cx + D (ax + bx + c) + + x x + 5 (x + ) dx x x + 5 (x + ) Ax + B x + + Cx + D (x + ) x x + 5 (x + ) (Ax + B)(x + ) + Cx + D (x + ) x x + 5 Ax + Bx + Ax + B + Cx + D x x + 5 Ax + Bx + (A + C)x + (B + D) A 0 B A + C B + D 5 C + D 5 D 4 x x + 5 (x + ) x + + x + 4 (x + ) x ( x + 5 (x + ) dx x + + x + 4 (x + ) x + dx x (x + ) dx + 4 Ex + F (ax + bx + c) n ) dx (x + ) dx u x + x tan θ du xdx dx sec θdθ du dx arctan x u du + 4 (tan θ + ) sec θdθ arctan x + u + 4 cos θdθ arctan x + (x + ) + ( + cos θ)dθ arctan x + [θ (x + ) + + ] sin θ + C arctan x + (x + [θ + sin θ cos θ] + C + ) 0
arctan x + (x + ) + arctan x + x + x + + C arctan x + (x + ) + x + x + + C