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FP2/3 Integration maths help!

RockConcert

Integral of sqrt of (1+t^2), using the substitution t=sinhx ?

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Reply 1

DylanJ42

Original post by RockConcert

Integral of sqrt of (1+t^2), using the substitution t=sinhx ?

just think of it as a normal C4 t=sinx sub except be careful with the identities (you know how the hyperbolic identities differ slightly from the trig ones)

Reply 2

RockConcert

Original post by DylanJ42

just think of it as a normal C4 t=sinx sub except be careful with the identities (you know how the hyperbolic identities differ slightly from the trig ones)

I keep trying with that approach but my answer is far too simple.. ( arsinht instead of 1/2(t(sqrt of (1+t^2))+arsinht) )

Reply 3

DylanJ42

Original post by RockConcert

I keep trying with that approach but my answer is far too simple.. ( arsinht instead of 1/2(t(sqrt of (1+t^2))+arsinht) )

maybe im wrong, let me try it, 2 secs

Reply 4

Bobjim12

integral becomes 1 + sinh^2(x)/cosh(x)

change 1 + sinh^2(x)

Reply 5

Zacken

Original post by RockConcert

I keep trying with that approach but my answer is far too simple.. ( arsinht instead of 1/2(t(sqrt of (1+t^2))+arsinht) )

Show us your working. We can't help until you do.

Oh, and moved to maths.

Reply 6

DylanJ42

Original post by RockConcert

I keep trying with that approach but my answer is far too simple.. ( arsinht instead of 1/2(t(sqrt of (1+t^2))+arsinht) )

okay got it, can you post some workings?

Reply 7

EricPiphany

Original post by RockConcert

I keep trying with that approach but my answer is far too simple.. ( arsinht instead of 1/2(t(sqrt of (1+t^2))+arsinht) )

You put the differential in 'upside down' :smile:

Reply 8

Zacken

Original post by RockConcert

As above, it should be

\int \cosh x \sqrt{1 + \sinh^2 x} \, \mathrm{d}x

Reply 9

drandy76

Dt=cosh(X)dx is what you needed

Posted from TSR Mobile

Reply 10

Louisb19

Original post by RockConcert

You got confused with your substitution. We want :

\mathrm{d} t = \mathrm{d}x \cosh{x}

And the answer will follow using the hyperbolic equivalent of

\sin{2\theta} = 2\cos{\theta} \sin{\theta}

You should have recognised your mistake right away as

\sqrt{1+t^2} \neq \dfrac{1}{\sqrt{1+t^2}}

(edited 8 years ago)

Reply 11

Zacken

Original post by Louisb19

And the answer will follow using the hyperbolic equivalent of

\sin{2\theta} = 2\cos{\theta} \sin{\theta}

Surely you mean

\cos 2\theta = 2\cos^2 \theta - 1

Reply 12

DylanJ42

Original post by Zacken

Surely you mean

\cos 2\theta = 2\cos^2 \theta - 1

i think he means after the integration has been done, using that identity to get the answer into the correct form

Reply 13

B_9710

\displaystyle \cosh 2x=2\cosh^2 x -1

(edited 8 years ago)

Reply 14

Zacken

Original post by DylanJ42

i think he means after the integration has been done, using that identity to get the answer into the correct form

Ah, okay.

Reply 15

EricPiphany

Original post by DylanJ42

i think he means after the integration has been done, using that identity to get the answer into the correct form

true dat PRSOM

Reply 16

RockConcert

Original post by Louisb19

You got confused with your substitution. We want :

\mathrm{d} t = \mathrm{d}x \cosh{x}

And the answer will follow using the hyperbolic equivalent of

\sin{2\theta} = 2\cos{\theta} \sin{\theta}

You should have recognised your mistake right away as

\sqrt{1+t^2} \neq \dfrac{1}{\sqrt{1+t^2}}

Ah thank you so much, that identity (sin2x) was the key! Man I'm not going to forget that identity again..

Reply 17

EricPiphany

Original post by RockConcert

Ah thank you so much, that identity (sin2x) was the key! Man I'm not going to forget that identity again..

remember it with the h on the end too. cosh

Reply 18

RockConcert

Original post by EricPiphany

remember it with the h on the end too. cosh

Don't worry that was a typo, I did use sinh2x = 2sinhxcoshx in my solution. But thank you for your concern!

Reply 19

EricPiphany

Original post by RockConcert

Don't worry that was a typo, I did use sinh2x = 2sinhxcoshx in my solution. But thank you for your concern!

You're welcome xxx

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