The Student Room Group

Integration by substitution

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

The answer is

(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
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

The answer is

(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?
Reply 2
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.
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?
Reply 4
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

As said before, any help and advice gratefully received..
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

As said before, any help and advice gratefully received..


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).
(edited 1 year ago)
Reply 6
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

As said before, any help and advice gratefully received..

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?
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

As said before, any help and advice gratefully received..



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.
Reply 8
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.

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