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Integration by sub

Hi

I have an integration by sub question

Use the sub x = 3cosh(u)

to show that

Integral of dx/Sqr root(X^2 -9)

= cosh^-1(x/3) + C

Do I sub in the value for x and start from there or do I differentiate the sub they have given? or make u the subject? Just a bit unsure how to start it off

Thanks for any help!
Original post by RHCPfan
Hi

I have an integration by sub question

Use the sub x = 3cosh(u)

to show that

Integral of dx/Sqr root(X^2 -9)

= cosh^-1(x/3) + C

Do I sub in the value for x and start from there or do I differentiate the sub they have given? or make u the subject? Just a bit unsure how to start it off

Thanks for any help!


Remember that they have dx and you'll be integrating with respect to u.. so what you've said about differentiating the substitution is correct, then think about how to get du in there rather than dx. :smile:
Reply 2
Avatar for TeeEm
TeeEm
19
Original post by RHCPfan
Hi

I have an integration by sub question

Use the sub x = 3cosh(u)

to show that

Integral of dx/Sqr root(X^2 -9)

= cosh^-1(x/3) + C

Do I sub in the value for x and start from there or do I differentiate the sub they have given? or make u the subject? Just a bit unsure how to start it off

Thanks for any help!


substitute directly

differentiate the substitution as it is given to get an expression for your dx
Reply 3
Avatar for RHCPfan
RHCPfan
OP
12
Thanks guys!
So I have got du/dx = 3sinhu
1/3sinhu = dx

and now the equation as

Integral of (1/Sqr root(3coshu)^2 - 9 ) * 1/3sinhu

Is that on the right track?
Original post by RHCPfan
Thanks guys!
So I have got du/dx = 3sinhu
1/3sinhu = dx

and now the equation as

Integral of (1/Sqr root(3coshu)^2 - 9 ) * 1/3sinhu

Is that on the right track?


Correct - now think of a relevant trig identity.

If you quote people they will be able to see that you've replied, like so. :smile:
Reply 5
Avatar for RHCPfan
RHCPfan
OP
12
Original post by SeanFM
Correct - now think of a relevant trig identity.

If you quote people they will be able to see that you've replied, like so. :smile:


Ah yes my bad there. Been on here long enough to know to do that but don't log on much any more. Thanks for your help by the way, got there in the end :smile: I just over complicated in my head :frown:
Original post by RHCPfan
Ah yes my bad there. Been on here long enough to know to do that but don't log on much any more. Thanks for your help by the way, got there in the end :smile: I just over complicated in my head :frown:


Great, I am glad that it is sorted. Well done :borat:
remember the hyperbolic identities are nearly the same as the ordinary ones... but with the odd minus thrown in.

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