Integration problem

1. log x^2

∫[log x^2]dx
= ∫[2*log x]dx
= ∫[2*log x]dx
= 2*∫(ln x/ln 10)dx
= (2/ln 10)*∫(ln x)dx
= (2/ln 10)*[x ln x - ∫x d(ln x)]
(integration by part)
= (2/ln 10)*[x ln x - ∫dx]
= (2/ln 10)*[x ln x - ∫dx]
= (2/ln 10)*[x ln x - x] + C

where C is arbitrary constant


2. (log x)^2
= ∫(log x)^2 dx
= x.(log x)^2 - ∫x d(log x)^2
integration by part)

= x.(log x)^2
- (1/ln 10).∫2.x.(log x).(1/x)dx
= x.(log x)^2 - (1/ln 10).∫2.(log x)dx
= x.(log x)^2 - (1/ln 10).∫2.(log x)dx
= x.(log x)^2
- (1/ln 10).(2/ln 10)*[x ln x - x]
(by question 1)
= x.(log x)^2 - 2/(ln 10)^2*[x ln x - x]
+ C (by question 1)

where C is arbitrary constant




2009-11-29 15:37:25 補充：
Sorry, I have forgotton to answer Q3). As below:
3. (sin(pi/x))/x^2 from 1 to 2

∫(-1 to 2) (sin(pi/x))/x^2 dx
= - ∫(-1 to 2) (sin(pi/x))d(1/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)

2009-11-29 15:37:53 補充：
Sorry, I have forgotton to answer Q3). As below:
3. (sin(pi/x))/x^2 from 1 to 2

∫(-1 to 2) (sin(pi/x))/x^2 dx
= - ∫(-1 to 2) (sin(pi/x))d(1/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)

2009-11-29 15:38:06 補充：
Let u = pi/x
x = -1, u = -pi
x = 2, u = pi / 2

So, the integral is :
= -(1/pi) ∫(-pi to pi/2) (sin u)du
= (1/pi) [(-pi to pi/2) cos u]
= (1/pi) [cos(-pi) - cos(pi/2)]
= (1/pi) [-1 - 0]
= -1/pi

2009-11-29 16:41:05 補充：
Sorry, I have forgotton to answer Q3). As below:
3. (sin(pi/x))/x^2 from 1 to 2

∫(-1 to 2) (sin(pi/x))/x^2 dx
= - ∫(-1 to 2) (sin(pi/x))d(1/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)

2009-11-29 16:41:26 補充：
Sorry, I have forgotton to answer Q3). As below:
3. (sin(pi/x))/x^2 from 1 to 2

∫(-1 to 2) (sin(pi/x))/x^2 dx
= - ∫(-1 to 2) (sin(pi/x))d(1/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)

2009-11-29 16:43:26 補充：
Sorry, I have forgotton to answer Q3). As below:
3. (sin(pi/x))/x^2 from 1 to 2

∫(-1 to 2) (sin(pi/x))/x^2 dx
= - ∫(-1 to 2) (sin(pi/x))d(1/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)

2009-11-29 16:43:39 補充：
......

2009-11-29 16:57:27 補充：
Testing

2009-11-29 18:12:30 補充：
Sorry, I have forgotton to answer Q3). As below:
3. (sin(pi/x))/x^2 from 1 to 2

∫(-1 to 2) (sin(pi/x))/x^2 dx
= - ∫(-1 to 2) (sin(pi/x))d(1/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)
= -(1/pi) ∫(-1 to 2) (sin(pi/x))d(pi/x)