Integration by Parts I. Basic Technique By the Product Rule for Derivatives, d f ( x) g ( x) f ( x) g ( x) g ( x) f ( x) . Thus, dx f (x)g (x) g(x) f (x) dx f (x)g(x) f (x)g (x) dx g(x) f (x) dx f ( x) g ( x) f ( x) g ( x) dx f ( x) g ( x) g ( x) f ( x) dx . This formula for integration by parts often makes it possible to reduce a complicated integral involving a product to a simpler integral. By letting u f ( x) du f ( x)dx dv g ( x)dx v g ( x) we get the more common formula for integration by parts: udv uv vdu Example 1: Find x ln xdx . 1 dx and v x 1 2 x . Thus, 2 1 x ln xdx (ln x)( x dx) udv uv vdu (ln x) x 2 2 1 1 11 1 2 1 1 2 x dx x 2 ln x x 2 C x dx x ln x 2 2 22 2 x 2 1 2 1 x ln x x 2 C . 2 4 Let u ln x and dv xdx du x dx It is possible that when you set up an integral using integration by parts, the resulting integral will be more complicated than the original integral. In this case, change your substitutions for u and dv. 1 Example 2: Find x sin x dx . 1 2 x . Thus, 2 1 udv uv vdu (sin x) x 2 2 Let u sin x and dv x dx du cos x dx and v x dx x sin x dx (sin x)( x dx) 1 1 2 x (cos x dx) 2 x cos x dx . Notice that this resulting integral is 2 2 more complicated that the original one. Therefore, let u x and dv sin x dx cos x . Thus, x sin x dx = udv uv vdu (x)( cos x) ( cos x) dx x cos x cos xdx sin x dx du dx and v x cos x sin x C . Example 3: Find ln x dx . 1 dx and v 1dx x . Thus, x 1 udv uv vdu (ln x)( x) dx ( x) x Let u ln x and dv 1dx du ln x dx (ln x)(1dx) x ln x 1dx x ln x x C . Example 4: Find arctan x dx . As in the previous example, let u arctan x and dv dx du arctan x dx udv uv vdu x arctan x x 1 2x dx x arctan x dx x arctan x dx 1 x 2 1 x and v 1dx x . Thus, 1 x 2 1 x 1 dx 1 x2 2 1 1 x arctan x ln x 2 1 C = x arctan x ln( x 2 1) C . 2 2 2 2 Example 5: Find ln x dx . x2 1 1 dx x 2 dx du dx and v x 2 dx 1x 1 2 x x 1 ln x 1 1 dx (ln x) 2 dx udv uv vdu (ln x) . Thus, 2 x x x x 1 ln x ln x 1 1 1 ln x dx x 2 dx C dx 2 x x x x x x x ln x 1 C. x Let u ln x and dv Sometimes, it is necessary to use integration by parts more than once. Example 6: Find x 2 cos x dx . Let u x 2 and dv cos x dx du 2 xdx and v cos x dx sin x . Thus, x cos x dx udv uv vdu (x )(sin x) (2xdx)(sin x) x sin x 2 x sin xdx . Notice that the resulting integral x sin x dx is less complicated 2 2 2 than the original one, but integration by parts is needed to evaluate it. Let u x and dv sin x dx du dx and v sin x dx cos x . Thus, x sin x dx udv uv vdu x( cos x) ( cos x) dx x cos x cos x dx x cos x sin x . Finally, we get x cos x dx x sin x 2 2 2 x cos x sin x C x 2 sin x 2 x cos x 2 sin x C . [Note: We will be able to integrate x cos x dx more easily with tabular integration; this 2 technique will be described later.] The next illustration of repeated integration by parts deserves special attention. Example 7: Find e sin x dx . x 3 sin xdx cos x . Thus, e sin x dx udv uv vdu (e )( cos x) ( cos x)(e dx) e cos x e cos xdx . Notice that integration by parts is now needed to evaluate e cos xdx . Let u e and dv cos xdx du e dx and v cos xdx sin x . Thus, e cos xdx udv uv vdu e sin x (sin x)(e dx) e sin x e sin xdx . Returning to the original problem, e sin x dx = e cos x e cos xdx e cos x e sin x e sin x dx . 1 Thus, 2 e sin x dx e cos x e sin x e sin x dx = e cos x 2 Let u e x and dv sin x dx du e x dx and v x x x x x x x x x x x x x x x x x x x x x x x x 1 x e sin x C and this does check. 2 The next example illustrates an interesting type of integral that surprisingly requires integration by parts. Example 8: sin x dx . Let u x u 2 x 2udu dx sin x dx (sin u)(2udu) x sin x dx = x cos x sin x C or u sin u du = u cos u sin u C . Thus, sin x dx = 2 u sin u du = 2u cos u 2 sin u C 2 x cos x 2 u sin u du . In example 2, we got the following using integration by parts: 2 sin x C and this does check. In general, to evaluate f x dx = n u n n 1 f x dx , let u n f (u ) du . 4 n x u n x nu n1du dx II. Tabular Integration f (x)g(x) dx , in which f can be differentiated repeatedly Integrals of the form to become zero and g can be integrated repeatedly without difficulty, can be evaluated using tabular integration. Example: Find xe 3 2x dx . f (x ) and its derivatives g (x ) and its antiderivatives x3 (+) e2x 3x 2 (–) 1 2x e 2 6x (+) 1 2x e 4 6 (–) 1 2x e 8 1 2x e 16 0 1 3 2x 3 2 2x 6 2x 6 2x x e x e xe e C 2 4 8 16 1 3 3 3 = x 3 e 2 x x 2 e 2 x xe2 x e 2 x C 2 4 4 8 x 3 e 2 x dx = III. Reduction Formulas Integration by parts can be used to derive reduction formulas for integrals. These are formulas that express an integral involving a power of a function in terms of an integral that involves a lower power of that function 5 Example 1: Prove the reduction formula use the result to find (ln x) dx x(ln x) n (ln x) n n n 1 dx and (ln x) 2 dx . 1 Let u (ln x) n and dv dx du n(ln x) n 1 dx and v 1dx x . Thus, x 1 (ln x) n dx udv uv vdu x(ln x) n ( x)n(ln x) n 1 dx x(ln x) n x n (ln x) dx . Thus, (ln x) dx x(ln x) 2 ln x dx x(ln x) 1dx] x(ln x) 2x ln x 2x C and this checks. n 1 2 2 2 2[ x(ln x) 2 Example 2: Prove the reduction formula n 1 (sin x) n2 dx . n 1 (sin x) n dx (sin x) n1 cos x n Let u (sin x) n1 and dv sin x dx du (n 1)(sin x) n2 cos x dx and v = sin x dx cos x . Thus, (sin x) n dx (sin x) n1 (sin x dx) udv uv vdu (sin x) n1 ( cos x) ( cos x)( n 1)(sin x) n2 cos x dx (sin x) n1 cos x (n 1) (sin x) n2 cos 2 x dx (sin x) n1 cos x (n 1) (sin x) dx n (sin x) dx (sin x) cos x (n 1) (sin x) 1 n 1 (sin x) dx (sin x) cos x (sin x) dx . n n (n 1) (sin x) n2 (1 sin 2 x) dx (sin x) n1 cos x (n 1) (sin x) n2 dx n n n 1 n n 1 n2 6 n2 dx Practice Sheet for Integration by Parts (1) 3 x ln x dx = (2) Prove the following reduction formula: n x m (ln x) n1 dx for m 1 . m 1 (3) (4) cos x dx = (5) (6) x arcsec x dx = (7) e x cos x dx = (8) x 3 sin x 2 dx = (9) e x dx = (10) x 2 e 3 x dx = (11) x 2 sin x dx = arcsin x dx = ln x x3 dx = 7 x m (ln x) n dx 1 x m1 (ln x) n m 1 (12) (13) (14) 3 e x dx 3 e x 3x dx sin(ln x)dx Solution Key for Integrations by Parts (1) Let u ln x and dv x 1 3 1 3 4 1 dx and v x 3 dx x 3 x 4 1 3 4 3 4 3 1 3 4 3 3 x 3 dx x 3 x dx x 4 4 4 x 4 dx du 3 x ln x dx = 3 x 4 3 ln x 4 9 43 x C. 16 1 (2) Let u (ln x) n and dv x m dx du n(ln x) n 1 dx and v x m dx x 1 1 1 1 n x m 1 x m ln x dx x m 1 (ln x) n n(ln x) n 1 dx x m 1 m 1 m x x m 1 1 n x m1 (ln x) n x m (ln x) n1 dx . m x m 1 arcsin x dx = x arcsin x 1 x2 x 1 x 1 dx x arcsin x 2 1 (3) Let u arcsin x and dv dx du 2 x arcsin x 1 x 2 C. 8 dx and v 1dx x 2x 1 x2 dx cos x dx = 2 wcos w dw . Let u w and dv cos w dw du dw and v cos w dw sin w cos x dx = 2 wcos w dw = 2w(sin w) 2 (sin w) dw 2w sin w 2 sin w dw (4) Let w x w 2 x 2wdw dx 2w sin w 2 cos w C 2 x sin( x ) 2 cos( x ) C . 1 1 1 dx du dx and v x 3 dx x 2 3 x x 2 ln x 1 1 1 2 1 1 dx = x 2 ln x x dx x 2 ln x x 3 dx 3 2 2 x 2 2 x (5) Let u ln x and dv 1 2 1 ln x 1 x ln x x 2 C 2 C. 2 2 4 2x 4x (6) Let u arc sec x and dv x dx du x arcsec x dx = 1 2 1 x arc sec x 2 2 1 dx and v x dx x x2 1 x 1 1 2 dx x 2 arc sec x x 1 C . 2 2 2 x 1 (7) Let u cos x and dv e x dx du sin x dx and v e x dx e x e x cos x e x sin xdx . Use integration by parts to evaluate e x cos x e x sin x e x cos x dx e x sin x dx . Let u sin x and dv e x dx du cos x dx and v e x dx e x e x sin x e x cos x dx . Thus, 1 2 x 2 e x sin x dx = e cos x dx = e cos x e sin xdx = x x x e x cos x dx e x cos x e x sin x 2 e x cos x dx e x cos x e x sin x e x cos x dx 9 e cos x dx x ex cos x sin x C . 2 (8) Let w x 2 dw 2 x dx x 3 sin x 2 dx 1 2 x sin x 2x dx 2 2 1 wsin w dw . Let u w and dv sin w dw du dw and v 2 cos w 1 2 w sin w dw 1 w cos w 2 sin w dw 1 cos w dw w cos w 2 1 1 1 sin w C x 2 cos x 2 sin x 2 C . 2 2 2 (9) Let w x w 2 x 2w dw dx dv e w dw du dw and v we w e w e x dx = 2 we w dw . Let u w and e dw e we dw we e dw w w w w e x dx = 2 we w dw 2we w 2e w C 2 x e w x 2e (10) Use tabular integration: f (x) and its derivatives g (x) and its antiderivatives x2 e 3x (+) 1 3x e 3 2x (–) 1 3x e 9 2 (+) 0 1 3x e 27 Thus, x 2 e 3 x dx = 1 2 3x 2 3x 2 3x x e xe e C. 3 9 27 10 x C. (11) Use tabular integration: f (x) and its derivatives g (x) and its antiderivatives x2 sin x (+) cos x 2x (–) sin x 2 (+) cos x 0 Thus, x 2 sin x dx = x 2 cos x 2 x sin x 2 cos x C . (12) Let w 3 x w3 x 3w 2 dw dx w e dw can 3 e x dx 3 w 2 e w dw . 2 w integrated using tabular integration: f (x) and its derivatives g (x) and its antiderivatives w2 (+) 2w (–) 2 (+) ew ew ew ew 0 w 2 e w dw = w 2 e w 2we w 2e w C 33 x2 e 3 x 6 3 xe 3 x 3 3 6e x + C. 3 (13) Let w x w x 3w dw dx 3 e x dx 3 w 2 e w dw 2 In solution to problem #9 previously, e 3 3 x w2e w dw 3 we w dw . dx 3 w x we dw we e 11 w w w e 3 3 x x dx 3 we w dw 3we w 3e w C 3 3 x e 3 x 3e 3 x C . 1 (14) Let u sin(ln x) and dv dx du cos(ln x) dx and v 1dx x x sin(ln x)dx x sin(ln x) cos(ln x) dx . Use integration by parts to evaluate 1 cos(ln x) dx . Let u cos(ln x) and dv dx du sin(ln x) dx and x v 1dx x cos(ln x) dx = x cos(ln x) sin(ln x) dx . Thus, sin(ln x)dx x sin(ln x) cos(ln x) dx = x sin(ln x) x cos(ln x) sin(ln x) dx x sin(ln x) x cos(ln x) sin(ln x) dx 2 sin(ln x) dx x sin(ln x) x cos(ln x) sin(ln x)dx 1 1 x sin(ln x) x cos(ln x) C . 2 2 12