3 Integration By Parts Formula ∫∫udv = uv − vdu I. Guidelines for Selecting u and dv: (There are always exceptions, but these are generally helpful.) “L-I-A-T-E” Choose ‘u’ to be the function that comes first in this list: L: Logrithmic Function I: Inverse Trig Function A: Algebraic Function T: Trig Function E: Exponential Function Example A: ∫ x3 ln x dx *Since lnx is a logarithmic function and x3 is an algebraic function, let: u = lnx (L comes before A in LIATE) dv = x3 dx 1 du = dx x x 4 v = x 3dx = ∫ 4 ∫∫x3 ln xdx = uv − vdu x 4 x 4 1 = (ln x) − dx 4 ∫ 4 x x 4 1 = (ln x) − x 3dx 4 4 ∫ x 4 1 x 4 = (ln x) − + C 4 4 4 x 4 x 4 = (ln x) − + C ANSWER 4 16 www.rit.edu/asc Page 1 of 7 Example B: ∫sin x ln(cos x) dx u = ln(cosx) (Logarithmic Function) dv = sinx dx (Trig Function [L comes before T in LIATE]) 1 du = (−sin x) dx = − tan x dx cos x v = ∫sin x dx = − cos x ∫sin x ln(cos x) dx = uv − ∫ vdu = (ln(cos x))(−cos x) − ∫ (−cos x)(− tan x)dx sin x = −cos x ln(cos x) − (cos x) dx ∫ cos x = −cos x ln(cos x) − ∫sin x dx = −cos x ln(cos x) + cos x + C ANSWER Example C: ∫sin −1 x dx *At first it appears that integration by parts does not apply, but let: u = sin −1 x (Inverse Trig Function) dv = 1 dx (Algebraic Function) 1 du = dx 1− x 2 v = ∫1dx = x ∫∫sin −1 x dx = uv − vdu 1 = (sin −1 x)(x) − x dx ∫ 2 1− x ⎛ 1 ⎞ = x sin −1 x − ⎜− ⎟ (1− x 2 ) −1/ 2 (−2x) dx ⎝ 2 ⎠∫ 1 = x sin −1 x + (1− x 2 )1/ 2 (2) + C 2 = x sin −1 x + 1− x 2 + C ANSWER www.rit.edu/asc Page 2 of 7 II. Alternative General Guidelines for Choosing u and dv: A. Let dv be the most complicated portion of the integrand that can be “easily’ integrated. B. Let u be that portion of the integrand whose derivative du is a “simpler” function than u itself. Example: ∫ x3 4 − x 2 dx *Since both of these are algebraic functions, the LIATE Rule of Thumb is not helpful. Applying Part (A) of the alternative guidelines above, we see that x 4 − x 2 is the “most complicated part of the integrand that can easily be integrated.” Therefore: dv = x 4 − x 2 dx u = x 2 (remaining factor in integrand) du = 2x dx 1 v = x 4 − x 2 dx = − (−2x)(4 − x 2 )1/ 2 dx ∫∫2 ⎛ 1 ⎞⎛ 2 ⎞ 1 = ⎜− ⎟⎜ ⎟ (4 − x 2 )3 / 2 = − (4 − x 2 )3 / 2 ⎝ 2 ⎠⎝ 3 ⎠ 3 ∫∫x3 4 − x3 dx = uv − vdu ⎛ 1 ⎞ 1 = (x 2 )⎜− (4 − x 2 )3 / 2 ⎟ − − (4 − x 2 )3 / 2 (2x) dx ⎝ 3 ⎠ ∫ 3 − x 2 1 = (4 − x 2 )3 / 2 − (4 − x 2 )3 / 2 (−2x) dx 3 3 ∫ − x 2 1 ⎛ 2 ⎞ = (4 − x 2 )3 / 2 − (4 − x 2 )5 / 2 ⎜ ⎟ + C 3 3 ⎝ 5 ⎠ − x 2 2 = (4 − x 2 )3 / 2 − (4 − x 2 )5 / 2 + C Answer 3 15 www.rit.edu/asc Page 3 of 7 III. Using repeated Applications of Integration by Parts: Sometimes integration by parts must be repeated to obtain an answer. Note: DO NOT switch choices for u and dv in successive applications. Example: ∫ x 2 sin x dx u = x 2 (Algebraic Function) dv = sin x dx (Trig Function) du = 2x dx v = ∫sin x dx = −cos x ∫ x 2 sin x dx = uv − ∫ vdu = x 2 (−cos x) − ∫ − cos x 2x dx = −x 2 cos x + 2 ∫ x cos x dx Second application of integration by parts: u = x (Algebraic function) (Making “same” choices for u and dv) dv = cos x (Trig function) du = dx v = ∫ cos x dx = sin x ∫ x 2 sin x dx = −x 2 cos x + 2 [uv − ∫ vdu] = −x 2 cos x + 2 [x sin x − ∫sin x dx] = −x 2 cos x + 2 [x sin x + cos x + c] = −x 2 cos x + 2x sin x + 2cos x + c Answer www.rit.edu/asc Page 4 of 7 Note: After each application of integration by parts, watch for the appearance of a constant multiple of the original integral. Example: ∫ e x cos x dx u = cos x (Trig function) dv = e x dx (Exponential function) du = −sin x dx v = ∫ e x dx = e x ∫ e x cos x dx = uv − ∫ vdu = cos x e x − ∫ e x (−sin x) dx = cos x e x + ∫ e x sin x dx Second application of integration by parts: u = sin x (Trig function) (Making “same” choices for u and dv) dv = e x dx (Exponential function) du = cos x dx v = ∫ e x dx = e x ∫ e x cos x dx = e x cos x + (uv − ∫ vdu) ∫ e x cos x dx = e x cos x + sin x e x − ∫ e x cos x dx Note appearance of original integral on right side of equation. Move to left side and solve for integral as follows: 2∫ e x cos x dx = e x cos x + e x sin x + C 1 e x cos x dx = (e x cos x + e x sin x) + C Answer ∫ 2 www.rit.edu/asc Page 5 of 7 Practice Problems: 1. ∫3x e −x dx ln x 2. dx ∫ x 2 3. ∫ x 2 cos x dx 4. ∫ x sin x cos x dx 5. ∫ cos −1 x dx 6. ∫ (ln x) 2 dx 7. ∫ x3 9 − x 2 dx 8. ∫ e 2x sin x dx 9. ∫ x 2 x −1 dx 1 10. dx ∫ x(ln x)3 www.rit.edu/asc Page 6 of 7 Solutions: 1. − 3xe−x − 3e− x + C u = 3x dv = e−x dx ln x 1 2. − − + C u = ln x x x 1 dv = dx x 2 3. x 2 sin x + 2x cos x − 2sin x + C u = x 2 dv = cos x dx x cos 2x sin 2x sin 2x 4. − + + C note: = sin x cos x 4 8 2 u = x dv = sin 2x cos x dx 5. x cos −1 x − 1− x 2 + C u = cos−1 x dv = dx 6. x(ln x) 2 − 2x ln x + 2x + C u = (ln x) 2 dv = dx x 2 2 7. − (9 − x 2 )3 / 2 − (9 − x 2 )5 / 2 + C u = x 2 3 15 dv = (4 − x 2 )1/ 2 x dx 2e 2x sin x e 2x cos x 8. − + C u = sin x 5 5 dv = e 2x dx 2x 2 (x −1)3 / 2 8x(x −1)5 / 2 16(x −1)7 / 2 9. − + + C u = x 2 3 15 105 dv = (x −1)1/ 2 dx −1 1 10. + C u = = (ln x) −3 2(ln x) 2 (ln x)3 1 dv = dx x www.rit.edu/asc Page 7 of 7 .