# Integration by substitution

#1
The question to find the integral of

1/((x^2)-1) by using the substitution x = sec Z and where the limits are 2 and 3

(8(3^0.5) - 9(2^0.5))/12

I have been going round in circles on this question and any help would be greatly appreciated
0
4 weeks ago
#2
(Original post by apsley)
The question to find the integral of

1/((x^2)-1) by using the substitution x = sec Z and where the limits are 2 and 3

(8(3^0.5) - 9(2^0.5))/12

I have been going round in circles on this question and any help would be greatly appreciated
Dont think that is the answer. But what do you get when you do the usual substitution?
0
#3
(Original post by mqb2766)
Dont think that is the answer. But what do you get when you do the usual substitution?
Oops. You are right. It should be the integral of

1/(((x^2)-1)3/2) by using the substitution x = sec Z and where the limits are 2 and 3.
0
4 weeks ago
#4
(Original post by apsley)
Oops. You are right. It should be the integral of

1/(((x^2)-1)3/2) by using the substitution x = sec Z and where the limits are 2 and 3.
and what did you try?
0
#5
(Original post by mqb2766)
and what did you try?
Well, I got this far..

Part A

If x = sec z then x = 1/cos z

therefore 1/(x^2 -1) = 1/((1/(cos z)^2 -1) =

1/((1 -(cos z) ^2) / cos z ^2)= 1/(sin z^2 )/ (cos z ^2) =

(cos z) ^2 / (sin z) ^2

therefore

1/((x^2 -1)) ^ 3/2 =

((cos z) ^ 3)/ ((sin z) ^ 3

Part B

Using the quotient rule

dx/dz = (v du/dz – u dv/dz )/ v ^2

where u = 1, du/dz = 0, v= cosz and dv/dx = - sin z

therefore

dx/dz = cos z * 0 – 1 *(-sin z) = sin z

Part C

therefore

(1/((x^2 -1)) ^ 3/2) dx=

((cos z) ^3) * sin z) ) / (sin z) ^3 ) =

((cos z) ^3) / ((sin z) ^2 ) =

cot z ^2 * sec z

0
4 weeks ago
#6
(Original post by apsley)
Well, I got this far..

Part A

If x = sec z then x = 1/cos z

therefore 1/(x^2 -1) = 1/((1/(cos z)^2 -1) =

1/((1 -(cos z) ^2) / cos z ^2)= 1/(sin z^2 )/ (cos z ^2) =

(cos z) ^2 / (sin z) ^2

therefore

1/((x^2 -1)) ^ 3/2 =

((cos z) ^ 3)/ ((sin z) ^ 3

Part B

Using the quotient rule

dx/dz = (v du/dz – u dv/dz )/ v ^2

where u = 1, du/dz = 0, v= cosz and dv/dx = - sin z

therefore

dx/dz = cos z * 0 – 1 *(-sin z) = sin z

Part C

therefore

(1/((x^2 -1)) ^ 3/2) dx=

((cos z) ^3) * sin z) ) / (sin z) ^3 ) =

((cos z) ^3) / ((sin z) ^2 ) =

cot z ^2 * sec z

Id keep it as sec(z) as they suggest and use corresponding pythagorean identity
tan^2(z) + 1 = sec^2(z)
Not that different from what you did, but more direct.
Then
dx/dz = ...
which is a standard derivative for sec(), so the integration problem becomes ...

Note it looks about right (your last two lines dont match) what you have at the end
cos(z)/sin^2(z)
You could try another substitution u=sin(z).
Last edited by mqb2766; 4 weeks ago
0
#7
(Original post by mqb2766)
and what did you try?
Well, I got this far..

Part A

If x = sec z then x = 1/cos z

therefore 1/(x^2 -1) = 1/((1/(cos z)^2 -1) =

1/((1 -(cos z) ^2) / cos z ^2)= 1/(sin z^2 )/ (cos z ^2) =

(cos z) ^2 / (sin z) ^2

therefore

1/((x^2 -1)) ^ 3/2 =

((cos z) ^ 3)/ ((sin z) ^ 3

Part B

Using the quotient rule

dx/dz = (v du/dz – u dv/dz )/ v ^2

where u = 1, du/dz = 0, v= cosz and dv/dx = - sin z

therefore

dx/dz = cos z * 0 – 1 *(-sin z) = sin z

Part C

therefore

(1/((x^2 -1)) ^ 3/2) dx=

((cos z) ^3) * sin z) ) / (sin z) ^3 ) =

((cos z) ^3) / ((sin z) ^2 ) =

cot z ^2 * sec z

(Original post by mqb2766)
Id keep it as sec(z) as they suggest and use corresponding pythagorean identity
tan^2(z) + 1 = sec^2(z)
Not that different from what you did, but more direct.
Then
dx/dz = ...
which is a standard derivative for sec(), so the integration problem becomes ...

Note it looks about right (your last two lines dont match) what you have at the end
cos(z)/sin^2(z)
You could try another substitution u=sin(z).
Sorry, only half of my calculations have uploaded, but by using the substitution x = sec z,

I reckon that I am now after the integral of

((cot z)^2) * sec z

Am I on the right track?
0
4 weeks ago
#8
(Original post by apsley)
Well, I got this far..

Part A

If x = sec z then x = 1/cos z

therefore 1/(x^2 -1) = 1/((1/(cos z)^2 -1) =

1/((1 -(cos z) ^2) / cos z ^2)= 1/(sin z^2 )/ (cos z ^2) =

(cos z) ^2 / (sin z) ^2

therefore

1/((x^2 -1)) ^ 3/2 =

((cos z) ^ 3)/ ((sin z) ^ 3

Part B

Using the quotient rule

dx/dz = (v du/dz – u dv/dz )/ v ^2

where u = 1, du/dz = 0, v= cosz and dv/dx = - sin z

therefore

dx/dz = cos z * 0 – 1 *(-sin z) = sin z

Part C

therefore

(1/((x^2 -1)) ^ 3/2) dx=

((cos z) ^3) * sin z) ) / (sin z) ^3 ) =

((cos z) ^3) / ((sin z) ^2 ) =

cot z ^2 * sec z

Sorry, only half of my calculations have uploaded, but by using the substitution x = sec z,

I reckon that I am now after the integral of

((cot z)^2) * sec z

Am I on the right track?
Id really use the hints in the previous post. Part a) is correct but would be simpler if you used sec() and tan(). Part b) is wrong. The derivative of sec(z) is not sin(z), that is the derivative of -cos(z). The derivative of sec(z) is a standard one in your formula sheet.
0
#9
(Original post by mqb2766)
Id really use the hints in the previous post. Part a) is correct but would be simpler if you used sec() and tan(). Part b) is wrong. The derivative of sec(z) is not sin(z), that is the derivative of -cos(z). The derivative of sec(z) is a standard one in your formula sheet.
Ah, ta.

I will give that a shot.
0
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