Integration by substitution

[borchester]

I have been going round in circles on this and any help would be greatly appreciated


The answer is



Many thanks in advance

 

[topsquark]

Can you show us what circles you've been going around in? Please show us your work.

Hint: If [imath]x = sec( \theta )[/imath] then what is [imath]x^2 - 1[/imath]? What is dx?

-Dan

 

[borchester]

Can you show us what circles you've been going around in? Please show us your work.

Hint: If [imath]x = sec( \theta )[/imath] then what is [imath]x^2 - 1[/imath]? What is dx?

-Dan

Hi.

Thanks for the quick response. My knowledge of computers is less than that of the integral calculus, so apologies for the mess

if x = secΘ

then x^2 = (secΘ)^2

and x^2 -1 = (secΘ)^2 -1 = (1/(cosΘ^2)) -1 = (1-cosΘ^2)/(cosΘ ^2) =

(sinΘ^2)/(cosΘ^2)

Therefore

1/((x^2-1)^3/2) = cosΘ^2/sinΘ^2

***************

If x = sec Θ then x = 1/cosΘ

and using the quotient rule where dx/dΘ = (vdu/dΘ -udv/dΘ)/(v^2)

and v = cosΘ, u = 1, dv/dΘ =-sinΘ and du/dΘ = 0

Therefore

dx/dΘ = ( cosΘ x 0 - 1 x (-sinΘ))/ (cosΘ^2) = sinΘ/ cosΘ^2

therefore dx = (sinΘ / cosΘ^2)dΘ

++++++++++++++++++++

(cosΘ^2)/sin^Θ)x(sin^Θ/cosΘ^2) = 1/sinΘ = secΘ

That wasn't the answer I had before, but it seems doable

Can you guys check my workings?

Many thanks

 

[Deleted member 4993]

Hi.

Thanks for the quick response. My knowledge of computers is less than that of the integral calculus, so apologies for the mess

if x = secΘ

then x^2 = (secΘ)^2

and x^2 -1 = (secΘ)^2 -1 = (1/(cosΘ^2)) -1 = (1-cosΘ^2)/(cosΘ ^2) =

(sinΘ^2)/(cosΘ^2)

Therefore

1/((x^2-1)^3/2) = cosΘ^2/sinΘ^2

***************

If x = sec Θ then x = 1/cosΘ

and using the quotient rule where dx/dΘ = (vdu/dΘ -udv/dΘ)/(v^2)

and v = cosΘ, u = 1, dv/dΘ =-sinΘ and du/dΘ = 0

Therefore

dx/dΘ = ( cosΘ x 0 - 1 x (-sinΘ))/ (cosΘ^2) = sinΘ/ cosΘ^2

therefore dx = (sinΘ / cosΘ^2)dΘ

++++++++++++++++++++

(cosΘ^2)/sin^Θ)x(sin^Θ/cosΘ^2) = 1/sinΘ = secΘ

That wasn't the answer I had before, but it seems doable

Can you guys check my workings?

Many thanks

1/((x^2-1)^3/2) = cosΘ^2/sinΘ^2

Incorrect - What happened to ^3/2 ? Try again......

 

[borchester]

Lad, I have been working on this problem for a month and I was hoping for some help. If you can't be bothered then thanks, but no thanks

 

[Deleted member 4993]

Lad, I have been working on this problem for a month and I was hoping for some help. If you can't be bothered then thanks, but no thanks

Can you guys check my workings?

That is what we were doing -

Not upto your standards? Beggars cannot be choosers - I hope you know!

 

[Dr.Peterson]

if x = secΘ
...
1/((x^2-1)^3/2) = cosΘ^2/sinΘ^2
...
dx/dΘ = ( cosΘ x 0 - 1 x (-sinΘ))/ (cosΘ^2) = sinΘ/ cosΘ^2
therefore dx = (sinΘ / cosΘ^2)dΘ
Can you guys check my workings?

Lad, I have been working on this problem for a month and I was hoping for some help. If you can't be bothered then thanks, but no thanks

He did check your workings, just as you asked. He told you that 1/((x^2-1)^(3/2)) = cosΘ^2/sinΘ^2 is a little wrong. It should be (cosΘ^2/sinΘ^2)^(3/2).

Do you see that?

Take it from there, and you should get the answer.

 

[topsquark]

Hi.

Thanks for the quick response. My knowledge of computers is less than that of the integral calculus, so apologies for the mess

if x = secΘ

then x^2 = (secΘ)^2

and x^2 -1 = (secΘ)^2 -1 = (1/(cosΘ^2)) -1 = (1-cosΘ^2)/(cosΘ ^2) =

(sinΘ^2)/(cosΘ^2)

Therefore

1/((x^2-1)^3/2) = cosΘ^2/sinΘ^2

***************

If x = sec Θ then x = 1/cosΘ

and using the quotient rule where dx/dΘ = (vdu/dΘ -udv/dΘ)/(v^2)

and v = cosΘ, u = 1, dv/dΘ =-sinΘ and du/dΘ = 0

Therefore

dx/dΘ = ( cosΘ x 0 - 1 x (-sinΘ))/ (cosΘ^2) = sinΘ/ cosΘ^2

therefore dx = (sinΘ / cosΘ^2)dΘ

++++++++++++++++++++

(cosΘ^2)/sin^Θ)x(sin^Θ/cosΘ^2) = 1/sinΘ = secΘ

That wasn't the answer I had before, but it seems doable

Can you guys check my workings?

Many thanks

You are making this waaaay too difficult.

[imath]x = sec( \theta )[/imath]

So [imath]x^2 - 1 = sec^2( \theta ) - 1 = tan^2( \theta )[/imath]

and [imath]dx = sec( \theta ) ~ tan( \theta ) ~d \theta[/imath]

Your integral turns into
[math]\int \dfrac{dx}{(x^2 - 1)^{3/2}} = \int \dfrac{ sec( \theta ) ~ tan( \theta ) ~ d \theta }{(tan^2( \theta ))^{3/2} }[/math]
See what you can do with this.

-Dan

 

[borchester]

Happily, the internet is full of better and

He did check your workings, just as you asked. He told you that 1/((x^2-1)^(3/2)) = cosΘ^2/sinΘ^2 is a little wrong. It should be (cosΘ^2/sinΘ^2)^(3/2).

Do you see that?

Take it from there, and you should get the answer.

Many thanks, I will give it a shot.

You are making this waaaay too difficult.

[imath]x = sec( \theta )[/imath]

So [imath]x^2 - 1 = sec^2( \theta ) - 1 = tan^2( \theta )[/imath]

and [imath]dx = sec( \theta ) ~ tan( \theta ) ~d \theta[/imath]

Your integral turns into
[math]\int \dfrac{dx}{(x^2 - 1)^{3/2}} = \int \dfrac{ sec( \theta ) ~ tan( \theta ) ~ d \theta }{(tan^2( \theta ))^{3/2} }[/math]
See what you can do with this.

-Dan

Many thanks

You are making this waaaay too difficult.

[imath]x = sec( \theta )[/imath]

So [imath]x^2 - 1 = sec^2( \theta ) - 1 = tan^2( \theta )[/imath]

and [imath]dx = sec( \theta ) ~ tan( \theta ) ~d \theta[/imath]

Your integral turns into
[math]\int \dfrac{dx}{(x^2 - 1)^{3/2}} = \int \dfrac{ sec( \theta ) ~ tan( \theta ) ~ d \theta }{(tan^2( \theta ))^{3/2} }[/math]
See what you can do with this.

-Dan

And many thanks. I have a tendency to confuse sec with cosec, but this has put me straight