Section 5.6. Integration By Parts. MATH 126 (Section 5.6) Integration By Parts The University of Kansas 1 / 10

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1 Section 5.6 Integration By Parts MATH 126 (Section 5.6) Integration By Parts The University of Kansas 1 / 10

2 Integration By Parts Manipulating the Product Rule d dx (f (x) g(x)) = f (x) g (x) + f (x) g(x) f (x) g(x) = ( f (x) g (x) + g(x) f (x) ) dx f (x) g (x) dx = f (x) g(x) g(x) f (x) dx We do not obtain a rule which solves the integral of a product, but we obtain a method which potentially simplifies the integration. f (x) g (x) dx = f (x) g(x) g(x) f (x) dx MATH 126 (Section 5.6) Integration By Parts The University of Kansas 2 / 10

3 Integration By Parts u dv = u v v du The aim of Integration By Parts is to manipulate the integrand into an expression with an easily calculated antiderivative. View the integrand as a product, one term you will integrate and the other term you will differentiate. Example (1): x sin(x) dx u = x dv = sin(x) dx du = 1 dx v = dv = cos(x) u dv = u v v du x sin(x) dx = x ( cos(x)) cos(x) dx = x cos(x) + sin(x) + C MATH 126 (Section 5.6) Integration By Parts The University of Kansas 3 / 10

4 Example (1) Revisited: x sin(x) dx The choice of u = x and dv = sin(x) is not the only way to implement Integration By Parts. For instance, choosing u = sin(x) dv = x dx du = cos(x) dx v = dv = 1 2 x 2 x sin(x) dx = 1 2 x 2 sin(x) 1 2 x 2 cos(x) dx There is nothing wrong with this equality, but we do not know how to 1 integrate 2 x 2 cos(x) dx!! In fact, we do NOW know how to evaluate the integral. 1 2 x 2 cos(x) dx = 1 2 x 2 sin(x) x sin(x) dx = 1 2 x 2 sin(x) + x cos(x) sin(x) + C MATH 126 (Section 5.6) Integration By Parts The University of Kansas 4 / 10

5 Using Integration By Parts we can integrate functions that we know how to differentiate. Example (2): ln x dx u = ln x dv = 1 dx du = 1 x dx v = dv = x ln x dx = x ln x x 1 x dx = x ln x x + C There is no limit to the number of times the Integration By Parts identity can be used. Example (3): x 2 e x dx = x 2 ex 2xe x dx = x 2 e x 2xe x + 2e x dx = x 2 e x 2xe x + 2e x + C MATH 126 (Section 5.6) Integration By Parts The University of Kansas 5 / 10

6 Definite Integrals and Integration By Parts Definite integrals are evaluated using indefinite integrals b ( ) b f (x) dx = f (x) dx a Therefore, when using Integration By Parts, evaluate the indefinite integral and then use the Fundamental Theorem of Calculus. b a f (x) g (x) dx = (f (b)g(b) f (a)g(a)) 1 Example (4): arctan(x) dx 0 As arctan(x) dx = x arctan(x) 1 2 ln 1 + x 2 + C, a b a g(x) f (x) dx 1 0 arctan(x) dx = π ln(2) MATH 126 (Section 5.6) Integration By Parts The University of Kansas 6 / 10

7 Tabular Integration By Parts Sometimes a problem requires the use of the Integration By Parts identity recursively, that is, you want to apply the identity multiple times choosing the next u from the previous du and choosing the next dv from the previous v. Example (5): Use a Table to integrate x 3 cos(x) dx = +(x 3 )(sin(x)) (3x 2 )( cos(x)) + (6x)( sin(x)) (6)(cos(x)) + C MATH 126 (Section 5.6) Integration By Parts The University of Kansas 7 / 10

8 Some functions create a cycle when repeatedly integrated or differentiated. For example, e x, cos(x), and sin(x) create integration and differentiation cycles. When recursively applying Integration By Parts and an integrand repeats, stop using integration by parts and solve the equation for the original integral. Example (6): e x cos(x) dx Using u = e x and dv = cos(x) dx we obtain e x cos(x) dx = e x sin(x) e x sin(x) dx Using u = e x and dv = sin(x) dx we obtain e x cos(x) dx = e x sin(x) + e x cos(x) e x cos(x) dx Solving the equality for the original integral yields e x cos(x) dx = 1 2 ex (sin(x) + cos(x)) + C MATH 126 (Section 5.6) Integration By Parts The University of Kansas 8 / 10

9 Reduction Formulas Example (7): Suppose that n 2 is an integer. Verify the reduction formula sin n (x) dx = 1 n cos(x) sinn 1 (x) + n 1 n sin n 2 (x) dx Solution: Letting u = sin n 1 (x) and dv = sin(x) dx we obtain du = (n 1) sin n 2 (x) cos(x) dx and v = cos(x). sin n (x) dx = cos(x) sin n 1 (x) + (n 1) sin n 2 (x) cos 2 (x) dx Since cos 2 (x) = 1 sin 2 (x) we have = cos(x) sin n 1 (x) + (n 1) sin n 2 (x) dx (n 1) Solving the equality for sin n (x) dx sin n (x) dx verifies the reduction formula. MATH 126 (Section 5.6) Integration By Parts The University of Kansas 9 / 10

10 Integration By Parts is a single tool available to you and can be used with mathematical tools to evaluate and reduce integrals. Example (8): π 0 e cos(t) sin(2t) dt An important identity that you should memorize this semester is sin(2t) = 2 sin(t) cos(t) Use Substitution with x = cos(t) to reduce the problem π 0 e cos(t) 2 sin(t) cos(t) dt = xe x dx We use Integration By Parts with u = x and dv = e x dx ( ) 1 = xe x e x = 2 1 e MATH 126 (Section 5.6) Integration By Parts The University of Kansas 10 / 10