Mar 10, Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 1 / 18

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1 Calculus with Algebra and Trigonometry II Lecture 14 Undoing the product rule: integration by parts Mar 10, 2015 Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 1 / 18

2 Undoing the product rule The product rule is (f (x)g(x)) = f (x)g(x) + f (x).g (x) Integrating we have f (x)g(x) = f (x).g (x) dx + g(x).f (x) dx It can be rewritten in the form f (x)g (x) dx = f (x)g(x) g(x)f (x) dx This allows us to transform an integral we can t evaluate, to one we can, hopefully. Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 2 / 18

3 Example 1 x sin(2x) dx f (x) = x g (x) = sin(2x) f (x) = 1 g(x) = 1 2 cos(2x) x sin(2x) dx = x ( 12 ) cos(2x) = 1 2 x cos(2x) ( 12 ) cos(2x) cos(2x) dx 1 dx = 1 2 x cos(2x) sin(2x) + C Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 3 / 18

4 Example 2 x 3 e x dx f (x) = x 3 g (x) = e x f (x) = 3x 2 g(x) = e x x 3 e x dx = x 3 ( e x ) = x 3 e x + 3 ( e x )(3x 2 dx) x 2 e x dx Now evaluate the integral by using by parts again Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 4 / 18

5 f (x) = x 2 g (x) = e x f (x) = 2x g(x) = e x x 3 e x dx = x 3 e x + 3 x 2 e x dx ( ) = x 3 e x + 3 x 2 e x ( e x )(2x dx) = ( x 3 3x 2 )e x + 6 xe x dx Integrating by parts again f (x) = x g (x) = e x f (x) = 1 g(x) = e x x 3 e x dx = ( x 3 3x 2 )e x + 6 xe x dx ( ) = ( x 3 3x 2 )e x + 6 xe x ( e x )1 dx = ( x 3 3x 2 6x 6)e x + C Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 5 / 18

6 Example 3 x 4 ln x dx f (x) = ln x g (x) = x 4 f (x) = 1 x g(x) = 1 5 x 5 ( ) ( ) ( ) x 4 ln x dx = 5 x 5 ln x 5 x 5 x dx = 1 5 x 5 ln x 1 x 4 dx 5 = 1 5 x 5 ln x 1 25 x 5 + C Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 6 / 18

7 Example 4 x sec 2 x dx f (x) = x g (x) = sec 2 x f (x) = 1 g(x) = tan x x sec 2 x dx = x tan x = x tan x (tan x)1 dx tan x dx = x tan x + ln cos x + C Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 7 / 18

8 Example 5 tan 1 x dx f (x) = tan 1 x g (x) = 1 f (x) = x 2 g(x) = x tan 1 x dx = x tan 1 x = x tan 1 x ( ) 1 x 1 + x 2 dx x 1 + x 2 dx We will use u substitution to evaluate the integral Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 8 / 18

9 u = 1 + x 2 du = 2x dx dx = du 2x ( ) x du tan 1 x dx = x tan 1 x u 2x = x tan 1 x u du = x tan 1 x 1 ln u + C 2 = x tan 1 x 1 2 ln(1 + x 2 ) + C Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 9 / 18

10 Example 6 cos x dx x = u 2 dx = 2u du cos x dx = 2 u cos du Now use integration by parts, let f (u) = u g (u) = cos u f (u) = 1 g(u) = sin u cos x dx = 2u sin u 2 sin u du = 2u sin u + 2 cos u + C = 2 x sin( x) + 2 cos( x) + C Calculus with Algebra and Trigonometry II Lecture 14Undoing Mar the product 10, 2015rule: integration 10 / 18

11 Example 7 cos 2 x dx f (x) = cos x g (x) = cos x f (x) = sin x g(x) = sin x cos 2 x dx = cos x sin x sin x( sin x dx) = cos x sin x + sin 2 x dx = cos x sin x + (1 cos 2 x) dx = cos x sin x + x cos 2 x dx cos 2 x dx = 1 (x + cos x sin x) + C 2 Calculus with Algebra and Trigonometry II Lecture 14Undoing Mar the product 10, 2015rule: integration 11 / 18

12 Example 8 1 x 2 dx f (x) = 1 x 2 g (x) = 1 f x (x) = 1 x 2 g(x) = x 1 x 2 dx = x 1 x 2 = x 1 x 2 + x x( dx) 1 x 2 x 2 1 x 2 dx = x x 2 1 x dx 1 x 2 Calculus with Algebra and Trigonometry II Lecture 14Undoing Mar the product 10, 2015rule: integration 12 / 18

13 1 x 2 dx = x 1 x 2 + x 2 1 dx + 1 x x 2 dx = x 1 x 2 1 x 2 dx + sin 1 x + C Rearranging gives 1 x 2 = 1 ( x ) 1 x sin 1 x + C Calculus with Algebra and Trigonometry II Lecture 14Undoing Mar the product 10, 2015rule: integration 13 / 18

14 Example 9 sec 3 x dx = sec x sec 2 x dx f (x) = sec x g (x) = sec 2 x f (x) = sec x tan x g(x) = tan x sec 3 x dx = sec x tan x tan x(sec x tan x dx) = sec x tan x secx tan 2 x dx = sec x tan x secx(sec 2 x 1) dx = sec x tan x sec 3 x dx + sec x dx sec 3 x = 1 (sec x tan x + ln sec x + tan x ) + C 2 Calculus with Algebra and Trigonometry II Lecture 14Undoing Mar the product 10, 2015rule: integration 14 / 18

15 Example 10 e ax sin bx dx f (x) = sin bx g (x) = e ax f (x) = b cos bx g(x) = 1 a eax e ax sin bx dx = 1 1 a eax sin bx a eax (b cos bx dx) = 1 a eax sin bx b e ax cos bx dx a Now use by parts to evaluate the integral Calculus with Algebra and Trigonometry II Lecture 14Undoing Mar the product 10, 2015rule: integration 15 / 18

16 f (x) = cos bx g (x) = e ax f (x) = b sin bxbx g(x) = 1 a eax e ax sin bx dx = 1 a eax sin bx b ( 1 1 a a eax cos bx a eax ( b sin bx dx)bx = 1 a eax sin bx b a 2 eax cos bx b2 a 2 e ax sin bx dx this implies ) (1 + b2 a 2 e ax sin bx dx = e ax ( 1 a sin bx b a 2 cos bx ) + C e ax sin bx dx = eax a 2 (a sin bx b cos bx) + C + b2 Calculus with Algebra and Trigonometry II Lecture 14Undoing Mar the product 10, 2015rule: integration 16 / 18

17 Example 10a e ax cos bx dx Assume it has the form e ax cos bx dx = Ae ax cos bx + Be ax sin bx + C Using the fundamental theorem we have e ax cos bx = (Ae ax cos bx = Be ax sin bx) = A(ae ax cos bx be ax sin bx) + B(e ax b cos bx + ae ax sin bx) = e ax cos bx(aa + bb) + e ax sin bx( ba + ab) Calculus with Algebra and Trigonometry II Lecture 14Undoing Mar the product 10, 2015rule: integration 17 / 18

18 This is true only if Solving we get this gives aa + bb = 1 ba + ab = 0 B = b a A A = and the integral becomes e ax cos bx dx = b2 aa + a A = 1 a a 2 + b 2 B = b a 2 + b 2 eax a 2 (b sin bx + a cos bx) + C + b2 Calculus with Algebra and Trigonometry II Lecture 14Undoing Mar the product 10, 2015rule: integration 18 / 18