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Integration Review Simplify the integrand if possible. – p. 1/12 Integration Review Simplify the integrand if possible. Look for an obvious or a clever substitution. – p. 1/12 Integration Review Simplify the integrand if possible. Look for an obvious or a clever substitution. Decide how you are going to integrate: * Integration by Parts * Trigonometric Integration * Trigonometric Substitution * Partial Fractions. – p. 1/12 Integration Review Simplify the integrand if possible. Look for an obvious or a clever substitution. Decide how you are going to integrate: * Integration by Parts * Trigonometric Integration * Trigonometric Substitution * Partial Fractions. If none of the above work, try to manipulate the integrand to transform it into an easier form which you can apply the above techniques. – p. 1/12 Integration Review : Simplify the Integrand if Possible. ! (x + sin x)2 dx – p. 2/12 Integration Review : Simplify the Integrand if Possible. ! (x + sin x)2 dx ! 2 = x + 2x sin x + sin2 xdx – p. 2/12 Integration Review : Simplify the Integrand if Possible. ! (x + sin x)2 dx ! 2 = x + 2x sin x + sin2 xdx ! 2 1 3 1. x dx = x + C 3 ! ! 2. 2x sin xdx = 2 x sin xdx ! = 2(−x cos x − (− cos x) · 1dx) by Integration by Parts, = −2x x + 2 sin x + C ! cos 3. sin2 xdx ! 1 − cos(2x) 1 1 = dx = (x − sin(2x)) + C 2 2 2 – p. 2/12 Integration Review : Simplify the Integrand if Possible. ! (x + sin x)2 dx ! 2 = x + 2x sin x + sin2 xdx ! 2 1 3 1. x dx = x + C 3 ! ! 2. 2x sin xdx = 2 x sin xdx ! = 2(−x cos x − (− cos x) · 1dx) by Integration by Parts, = −2x x + 2 sin x + C ! cos 3. sin2 xdx ! 1 − cos(2x) 1 1 = dx = (x − sin(2x)) + C 2 2 2 Therefore, ! 1 3 1 1 2 (x+sin x) dx = x −2x cos x+2 sin x+ x− sin(2x)+C 3 2 4 – p. 2/12 Integration Review : Simplify the integrand if Possible. ! sec x cos(2x) dx Try 2 sin x + sec x – p. 3/12 Integration Review: u-Substitution. ! 1 √ dx √ x+x x – p. 4/12 Integration Review: u-Substitution. ! 1 √ dx √ x+x x √ 1 Let u = x so that du = √ dx and x = u2 . 2 x – p. 4/12 Integration Review: u-Substitution. ! 1 √ dx √ x+x x √ 1 Let u = x so that du = √ dx and x = u2 . 2 x Then ! ! 1 1 √ dx = √ dx √ x+x x (1 + x) x – p. 4/12 Integration Review: u-Substitution. ! 1 √ dx √ x+x x √ 1 Let u = x so that du = √ dx and x = u2 . 2 x Then ! ! 1 1 √ dx = √ dx √ x+x x (1 + x) x ! ! 1 1 = 2du = 2 du 2 2 1+u 1+u – p. 4/12 Integration Review: u-Substitution. ! 1 √ dx √ x+x x √ 1 Let u = x so that du = √ dx and x = u2 . 2 x Then ! ! 1 1 √ dx = √ dx √ x+x x (1 + x) x ! ! 1 1 = 2du = 2 du 2 2 1+u 1+u √ −1 −1 = 2 tan u + C = 2 tan x+C – p. 4/12 Integration Review: u-Substitution. Try ! Try ! √ −1 tan xdx e √ x dx – p. 5/12 Integration Review: u-Subsitution and Trigonometric Substitution. ! 1 √ dx x2 4x2 + 1 – p. 6/12 Integration Review: u-Subsitution and Trigonometric Substitution. ! 1 √ dx x2 4x2 + 1 √ Try with u = 4x2 + 1 and complete the integration with a trigonometric substitution. – p. 6/12 Clever u-Substitution ! ln(tan x) dx sin x cos x – p. 7/12 Clever u-Substitution ! ln(tan x) dx sin x cos x Let u = ln(tan x) so that cos x 1 1 1 2 · sec xdx = · dx = dx. du = 2 tan x sin x cos x sin x cos x – p. 7/12 Clever u-Substitution ! ln(tan x) dx sin x cos x Let u = ln(tan x) so that cos x 1 1 1 2 · sec xdx = · dx = dx. du = 2 tan x sin x cos x sin x cos x Then ! ln(tan x) ! dx = udu sin x cos x – p. 7/12 Clever u-Substitution ! ln(tan x) dx sin x cos x Let u = ln(tan x) so that cos x 1 1 1 2 · sec xdx = · dx = dx. du = 2 tan x sin x cos x sin x cos x Then ! ln(tan x) ! dx = udu sin x cos x 1 1 2 = u + C = (ln(tan x))2 + C . 2 2 – p. 7/12 Clever u-Substitution. Try ! Try ! cos x cos3 (sin x)dx 1 1 + 2ex − e−x dx – p. 8/12 Integration by Parts ! ln(x + 1) dx 2 x – p. 9/12 Integration by Parts ! ln(x + 1) dx 2 x ! 1 1 1 = − ln(x + 1) − (− ) dx by Integration by Parts, x x x+1 – p. 9/12 Integration by Parts ! ln(x + 1) dx 2 x ! 1 1 1 = − ln(x + 1) − (− ) dx by Integration by Parts, x x x+1 ! 1 1 = − ln(x + 1) + dx x x(x + 1) – p. 9/12 Integration by Parts ! ln(x + 1) dx 2 x ! 1 1 1 = − ln(x + 1) − (− ) dx by Integration by Parts, x x x+1 ! 1 1 = − ln(x + 1) + dx x x(x + 1) ! 1 1 1 − dx by Partial Fractions, = − ln(x + 1) + x x x+1 – p. 9/12 Integration by Parts ! ln(x + 1) dx 2 x ! 1 1 1 = − ln(x + 1) − (− ) dx by Integration by Parts, x x x+1 ! 1 1 = − ln(x + 1) + dx x x(x + 1) ! 1 1 1 − dx by Partial Fractions, = − ln(x + 1) + x x x+1 1 = − ln(x + 1) + ln |x| − ln |x + 1| + C x – p. 9/12 Integration by Parts ! ln(x + 1) dx 2 x ! 1 1 1 = − ln(x + 1) − (− ) dx by Integration by Parts, x x x+1 ! 1 1 = − ln(x + 1) + dx x x(x + 1) ! 1 1 1 − dx by Partial Fractions, = − ln(x + 1) + x x x+1 1 = − ln(x + 1) + ln |x| − ln |x + 1| + C x x = ln +C 1 (x + 1) x (x + 1) – p. 9/12 Trigonometric Integration ! tan4 xdx – p. 10/12 Trigonometric Integration ! tan4 xdx ! = (tan2 x)2 dx – p. 10/12 Trigonometric Integration ! tan4 xdx ! = (tan2 x)2 dx ! = (sec2 x − 1)2 dx – p. 10/12 Trigonometric Integration ! tan4 xdx ! = (tan2 x)2 dx ! = (sec2 x − 1)2 dx ! = sec4 x − 2 sec2 x + 1dx – p. 10/12 Trigonometric Integration ! tan4 xdx ! = (tan2 x)2 dx ! = (sec2 x − 1)2 dx ! = sec4 x − 2 sec2 x + 1dx ! = (tan2 x + 1) sec2 x − 2 sec2 x + 1dx – p. 10/12 Trigonometric Integration ! tan4 xdx ! = (tan2 x)2 dx ! = (sec2 x − 1)2 dx ! = sec4 x − 2 sec2 x + 1dx ! = (tan2 x + 1) sec2 x − 2 sec2 x + 1dx 1 = tan3 x + tan x − 2 tan x + x + C 3 – p. 10/12 Trigonometric Integration ! tan4 xdx ! = (tan2 x)2 dx ! = (sec2 x − 1)2 dx ! = sec4 x − 2 sec2 x + 1dx ! = (tan2 x + 1) sec2 x − 2 sec2 x + 1dx 1 = tan3 x + tan x − 2 tan x + x + C 3 1 = tan3 x − tan x + x + C . 3 – p. 10/12 Trigonometric Integration Try !π 4 0 tan5 θ sec3 θdθ – p. 11/12 Partial Fractions Try ! 1 dx 4 x(x + 1) – p. 12/12