Page 1 13 Integration by Multiple Techniques Whenever we encounter a complex integral, often the rst thing to do is to use substitution to eliminate the composite function(s). Example: I = R x 3 e 1=x dx: Solution: Note that e 1=x is a composite fucntion. The substitution to change it into a simple, elementary function is u = 1=x, du = u 0 dx =x 2 dx or x 2 dx =du = d(1=x). Thus, I = Z x 3 e 1=x dx = Z (1=x)e 1=x d(1=x) = Z ue u du =[ue u e u ] + C = e 1=x [1 1=x] + C: Page 2 13 Integration by Multiple Techniques Whenever we encounter a complex integral, often the rst thing to do is to use substitution to eliminate the composite function(s). Example: I = R x 3 e 1=x dx: Solution: Note that e 1=x is a composite fucntion. The substitution to change it into a simple, elementary function is u = 1=x, du = u 0 dx =x 2 dx or x 2 dx =du = d(1=x). Thus, I = Z x 3 e 1=x dx = Z (1=x)e 1=x d(1=x) = Z ue u du =[ue u e u ] + C = e 1=x [1 1=x] + C: Example: I = R cos( p x)dx: Solution: Note that cos( p x) is a composite fucntion. The substitution to change it into a simple, elementary function is u = p x, du = u 0 dx = 1 2 p x dx or dx = 2 p xdu = 2udu. However, note that x = ( p x) 2 , I = Z cos( p x)dx = Z cos( p x)d( p x) 2 = Z cos(u)du 2 = 2 Z ucos(u)du = 2 Z udsin(u) = 2[usin(u) + cos(u)] + C = 2[ p xsin( p x) + cos( p x)] + C: Example: I = R sec(x)dx = R (1=cos(x))dx: Solution: This is a dicult integral. Note that 1=cos(x) is a composite fucntion. One substitu- tion to change it into a simple, elementary function is u = cos(x). However, x = cos 1 (u) and dx = x 0 du = du p 1u 2 . 1= p 1 u 2 is still a composite function that is dicult to deal with. Here is how we can deal with it. It involves trigonometric substitution followed by partial fractions. I = Z dx cos(x) = Z cos(x)dx cos 2 (x) = Z dsin(x) 1 sin 2 (x) = Z du 1 u 2 = Z [ 1 u 1 1 u + 1 ] du 2 = ln r j u + 1 u 1 j + C = lnj u + 1 p 1 u 2 j + C = lnj sin(x) + 1 cos(x) j + C = lnjtan(x) + sec(x)j + C; where u = sin(x) and p 1 u 2 = cos(x) were used! More Exercises on Integration by Multiple techniques: 1. Substitution followed by Integration by Parts Example: (i) Z x 3 sin(x 2 )dx = 1 2 Z x 2 sin(x 2 )dx 2 = 1 2 Z u sin udu = 1 2 Z ud cosu = 1 2 [sin u u cosu] + C = 1 2 [sin(x 2 ) x 2 cos(x 2 )] + C: Page 3 13 Integration by Multiple Techniques Whenever we encounter a complex integral, often the rst thing to do is to use substitution to eliminate the composite function(s). Example: I = R x 3 e 1=x dx: Solution: Note that e 1=x is a composite fucntion. The substitution to change it into a simple, elementary function is u = 1=x, du = u 0 dx =x 2 dx or x 2 dx =du = d(1=x). Thus, I = Z x 3 e 1=x dx = Z (1=x)e 1=x d(1=x) = Z ue u du =[ue u e u ] + C = e 1=x [1 1=x] + C: Example: I = R cos( p x)dx: Solution: Note that cos( p x) is a composite fucntion. The substitution to change it into a simple, elementary function is u = p x, du = u 0 dx = 1 2 p x dx or dx = 2 p xdu = 2udu. However, note that x = ( p x) 2 , I = Z cos( p x)dx = Z cos( p x)d( p x) 2 = Z cos(u)du 2 = 2 Z ucos(u)du = 2 Z udsin(u) = 2[usin(u) + cos(u)] + C = 2[ p xsin( p x) + cos( p x)] + C: Example: I = R sec(x)dx = R (1=cos(x))dx: Solution: This is a dicult integral. Note that 1=cos(x) is a composite fucntion. One substitu- tion to change it into a simple, elementary function is u = cos(x). However, x = cos 1 (u) and dx = x 0 du = du p 1u 2 . 1= p 1 u 2 is still a composite function that is dicult to deal with. Here is how we can deal with it. It involves trigonometric substitution followed by partial fractions. I = Z dx cos(x) = Z cos(x)dx cos 2 (x) = Z dsin(x) 1 sin 2 (x) = Z du 1 u 2 = Z [ 1 u 1 1 u + 1 ] du 2 = ln r j u + 1 u 1 j + C = lnj u + 1 p 1 u 2 j + C = lnj sin(x) + 1 cos(x) j + C = lnjtan(x) + sec(x)j + C; where u = sin(x) and p 1 u 2 = cos(x) were used! More Exercises on Integration by Multiple techniques: 1. Substitution followed by Integration by Parts Example: (i) Z x 3 sin(x 2 )dx = 1 2 Z x 2 sin(x 2 )dx 2 = 1 2 Z u sin udu = 1 2 Z ud cosu = 1 2 [sin u u cosu] + C = 1 2 [sin(x 2 ) x 2 cos(x 2 )] + C: (ii) Z xtan 1 p xdx = Z ( p x) 2 tan 1 p xd( p x) 2 = Z u 2 tan 1 udu 2 = 1 2 Z tan 1 udu 4 = 1 2 [u 4 tan 1 u Z u 4 dtan 1 u] = 1 2 [u 4 tan 1 u Z u 4 1 + 1 1 + u 2 du] = 1 2 [u 4 tan 1 u Z (u 2 1+ 1 1 + u 2 )du] = 1 2 [u 4 tan 1 u u 3 3 + u tan 1 u] + C = 1 2 [(x 2 1)tan 1 p x x 3=2 3 + p x] + C: Edwards/Penney 5 th ed 9.3 Problems. (15) Z x p x + 3dx; u = x + 3 (17) Z x 5 p x 3 + 1dx; u = x 3 + 1 (23) Z sec 1 p xdx; u = p x (29) Z x 3 cos(x 2 )dx; u = x 2 (25) Z tan 1 p xdx; u = p x (31) Z ln x x p x dx; u = p x (37) Z e p x dx; u = p x 2. Substitution followed by Integration by Partial Fractions Example: (i) Z cos xdx sin 2 x sin x 6 = Z d sin x sin 2 x sin x 6 = Z du (u 3)(u + 2) = 1 5 lnj u 3 u + 2 j+C = 1 5 lnj sin x 3 sin x + 2 j+C Edwards/Penney 5 th ed 9.5 Problems (difult ones!) (40) Z sec 2 t tan 3 t + tan 2 t dt (37) Z e 4t (e 2t 1) 3 dt (39) Z 1 + ln t t(3 + 2 ln t) 2 dt 2. Integration by Parts followed by Partial Fractions Page 4 13 Integration by Multiple Techniques Whenever we encounter a complex integral, often the rst thing to do is to use substitution to eliminate the composite function(s). Example: I = R x 3 e 1=x dx: Solution: Note that e 1=x is a composite fucntion. The substitution to change it into a simple, elementary function is u = 1=x, du = u 0 dx =x 2 dx or x 2 dx =du = d(1=x). Thus, I = Z x 3 e 1=x dx = Z (1=x)e 1=x d(1=x) = Z ue u du =[ue u e u ] + C = e 1=x [1 1=x] + C: Example: I = R cos( p x)dx: Solution: Note that cos( p x) is a composite fucntion. The substitution to change it into a simple, elementary function is u = p x, du = u 0 dx = 1 2 p x dx or dx = 2 p xdu = 2udu. However, note that x = ( p x) 2 , I = Z cos( p x)dx = Z cos( p x)d( p x) 2 = Z cos(u)du 2 = 2 Z ucos(u)du = 2 Z udsin(u) = 2[usin(u) + cos(u)] + C = 2[ p xsin( p x) + cos( p x)] + C: Example: I = R sec(x)dx = R (1=cos(x))dx: Solution: This is a dicult integral. Note that 1=cos(x) is a composite fucntion. One substitu- tion to change it into a simple, elementary function is u = cos(x). However, x = cos 1 (u) and dx = x 0 du = du p 1u 2 . 1= p 1 u 2 is still a composite function that is dicult to deal with. Here is how we can deal with it. It involves trigonometric substitution followed by partial fractions. I = Z dx cos(x) = Z cos(x)dx cos 2 (x) = Z dsin(x) 1 sin 2 (x) = Z du 1 u 2 = Z [ 1 u 1 1 u + 1 ] du 2 = ln r j u + 1 u 1 j + C = lnj u + 1 p 1 u 2 j + C = lnj sin(x) + 1 cos(x) j + C = lnjtan(x) + sec(x)j + C; where u = sin(x) and p 1 u 2 = cos(x) were used! More Exercises on Integration by Multiple techniques: 1. Substitution followed by Integration by Parts Example: (i) Z x 3 sin(x 2 )dx = 1 2 Z x 2 sin(x 2 )dx 2 = 1 2 Z u sin udu = 1 2 Z ud cosu = 1 2 [sin u u cosu] + C = 1 2 [sin(x 2 ) x 2 cos(x 2 )] + C: (ii) Z xtan 1 p xdx = Z ( p x) 2 tan 1 p xd( p x) 2 = Z u 2 tan 1 udu 2 = 1 2 Z tan 1 udu 4 = 1 2 [u 4 tan 1 u Z u 4 dtan 1 u] = 1 2 [u 4 tan 1 u Z u 4 1 + 1 1 + u 2 du] = 1 2 [u 4 tan 1 u Z (u 2 1+ 1 1 + u 2 )du] = 1 2 [u 4 tan 1 u u 3 3 + u tan 1 u] + C = 1 2 [(x 2 1)tan 1 p x x 3=2 3 + p x] + C: Edwards/Penney 5 th ed 9.3 Problems. (15) Z x p x + 3dx; u = x + 3 (17) Z x 5 p x 3 + 1dx; u = x 3 + 1 (23) Z sec 1 p xdx; u = p x (29) Z x 3 cos(x 2 )dx; u = x 2 (25) Z tan 1 p xdx; u = p x (31) Z ln x x p x dx; u = p x (37) Z e p x dx; u = p x 2. Substitution followed by Integration by Partial Fractions Example: (i) Z cos xdx sin 2 x sin x 6 = Z d sin x sin 2 x sin x 6 = Z du (u 3)(u + 2) = 1 5 lnj u 3 u + 2 j+C = 1 5 lnj sin x 3 sin x + 2 j+C Edwards/Penney 5 th ed 9.5 Problems (difult ones!) (40) Z sec 2 t tan 3 t + tan 2 t dt (37) Z e 4t (e 2t 1) 3 dt (39) Z 1 + ln t t(3 + 2 ln t) 2 dt 2. Integration by Parts followed by Partial Fractions Example: (i) Z (2x + 2) tan 1 xdx = Z tan 1 xd(x 2 + 2x) = (x 2 + 2x) tan 1 x Z (x 2 + 2x)d tan 1 x = (x 2 + 2x) tan 1 x Z 1 + x 2 + 2x 1 1 + x 2 dx = (x 2 + 2x) tan 1 x Z [dx + d(1 + x 2 ) 1 + x 2 dx 1 + x 2 ] = (x 2 + 2x) tan 1 x x ln(1 + x 2 ) + tan 1 x + C = (x + 1) 2 tan 1 x x ln(1 + x 2 ) + C: Li's Problems (difult ones!) (Li) Z (2t + 1) lntdt Page 5 13 Integration by Multiple Techniques Whenever we encounter a complex integral, often the rst thing to do is to use substitution to eliminate the composite function(s). Example: I = R x 3 e 1=x dx: Solution: Note that e 1=x is a composite fucntion. The substitution to change it into a simple, elementary function is u = 1=x, du = u 0 dx =x 2 dx or x 2 dx =du = d(1=x). Thus, I = Z x 3 e 1=x dx = Z (1=x)e 1=x d(1=x) = Z ue u du =[ue u e u ] + C = e 1=x [1 1=x] + C: Example: I = R cos( p x)dx: Solution: Note that cos( p x) is a composite fucntion. The substitution to change it into a simple, elementary function is u = p x, du = u 0 dx = 1 2 p x dx or dx = 2 p xdu = 2udu. However, note that x = ( p x) 2 , I = Z cos( p x)dx = Z cos( p x)d( p x) 2 = Z cos(u)du 2 = 2 Z ucos(u)du = 2 Z udsin(u) = 2[usin(u) + cos(u)] + C = 2[ p xsin( p x) + cos( p x)] + C: Example: I = R sec(x)dx = R (1=cos(x))dx: Solution: This is a dicult integral. Note that 1=cos(x) is a composite fucntion. One substitu- tion to change it into a simple, elementary function is u = cos(x). However, x = cos 1 (u) and dx = x 0 du = du p 1u 2 . 1= p 1 u 2 is still a composite function that is dicult to deal with. Here is how we can deal with it. It involves trigonometric substitution followed by partial fractions. I = Z dx cos(x) = Z cos(x)dx cos 2 (x) = Z dsin(x) 1 sin 2 (x) = Z du 1 u 2 = Z [ 1 u 1 1 u + 1 ] du 2 = ln r j u + 1 u 1 j + C = lnj u + 1 p 1 u 2 j + C = lnj sin(x) + 1 cos(x) j + C = lnjtan(x) + sec(x)j + C; where u = sin(x) and p 1 u 2 = cos(x) were used! More Exercises on Integration by Multiple techniques: 1. Substitution followed by Integration by Parts Example: (i) Z x 3 sin(x 2 )dx = 1 2 Z x 2 sin(x 2 )dx 2 = 1 2 Z u sin udu = 1 2 Z ud cosu = 1 2 [sin u u cosu] + C = 1 2 [sin(x 2 ) x 2 cos(x 2 )] + C: (ii) Z xtan 1 p xdx = Z ( p x) 2 tan 1 p xd( p x) 2 = Z u 2 tan 1 udu 2 = 1 2 Z tan 1 udu 4 = 1 2 [u 4 tan 1 u Z u 4 dtan 1 u] = 1 2 [u 4 tan 1 u Z u 4 1 + 1 1 + u 2 du] = 1 2 [u 4 tan 1 u Z (u 2 1+ 1 1 + u 2 )du] = 1 2 [u 4 tan 1 u u 3 3 + u tan 1 u] + C = 1 2 [(x 2 1)tan 1 p x x 3=2 3 + p x] + C: Edwards/Penney 5 th ed 9.3 Problems. (15) Z x p x + 3dx; u = x + 3 (17) Z x 5 p x 3 + 1dx; u = x 3 + 1 (23) Z sec 1 p xdx; u = p x (29) Z x 3 cos(x 2 )dx; u = x 2 (25) Z tan 1 p xdx; u = p x (31) Z ln x x p x dx; u = p x (37) Z e p x dx; u = p x 2. Substitution followed by Integration by Partial Fractions Example: (i) Z cos xdx sin 2 x sin x 6 = Z d sin x sin 2 x sin x 6 = Z du (u 3)(u + 2) = 1 5 lnj u 3 u + 2 j+C = 1 5 lnj sin x 3 sin x + 2 j+C Edwards/Penney 5 th ed 9.5 Problems (difult ones!) (40) Z sec 2 t tan 3 t + tan 2 t dt (37) Z e 4t (e 2t 1) 3 dt (39) Z 1 + ln t t(3 + 2 ln t) 2 dt 2. Integration by Parts followed by Partial Fractions Example: (i) Z (2x + 2) tan 1 xdx = Z tan 1 xd(x 2 + 2x) = (x 2 + 2x) tan 1 x Z (x 2 + 2x)d tan 1 x = (x 2 + 2x) tan 1 x Z 1 + x 2 + 2x 1 1 + x 2 dx = (x 2 + 2x) tan 1 x Z [dx + d(1 + x 2 ) 1 + x 2 dx 1 + x 2 ] = (x 2 + 2x) tan 1 x x ln(1 + x 2 ) + tan 1 x + C = (x + 1) 2 tan 1 x x ln(1 + x 2 ) + C: Li's Problems (difult ones!) (Li) Z (2t + 1) lntdt 14 Applications of Integrals The central idea in all sorts of application problems involving integrals is breaking the whole into pieces and then adding the pieces up to obtain the whole. 14.1 Area under a curve. We know the area of a rectangle is: Area = height width; (see Fig. 1(a)). This formula can not be applied to the area under the curve in Fig. 1(b) because the top side of the area is NOT a straight, horizontal line. However, if we divide the area into innitely many rectangles with innitely thin width dx, the innitely small area of the rectangle located at x is dA = height infintely small width = f(x)dx: Adding up the area of all the thin rectangles, we obtain the area under the curve Area = A(b) A(a) = A(b) Z 0 dA = b Z a f(x)dx; where A(x) = R x a f(s)ds is the area under the curve between a and x. 14.2 Volume of a solid. We know the volume of a cylinder is V olume = (cross section area) length; (see Fig. 2(a)). This formula does not apply to the volume of the solid obtained by rotating a curve around the x-axis (see Fig. 2(b)). This is because the cross-section area is NOT aRead More

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