Year 12s: what will be the first way you research your future options? >>

FP3 Integration

Trying to integrate

\frac{1}{\sqrt{x^2+4x+5}}

. My book, and Wolfram give the answer

arsinh(x+2)

, using a hyperbolic substitution, but I get the answer

ln|sec\theta +tan\theta|=ln|x^2+5x+7|

using a trigonometric substitution (

x + 2 = tan\theta

). Am I wrong, or are there just multiple ways of doing this?

Reply 1

olipal

Hmm, I guess for a question like this I would probably play it safe and just complete the square in the denominator. Unless they specifically ask for a substitution, that's normally the best move.

Reply 2

ViralRiver

They did ask for a substitution, but didn't specify whether it had to be hyperbolic or trigonometric.

Reply 3

olipal

Neither, in this case it would be to let u=x+2.

Reply 4

mir3a

Complete the square, and let u=x+2. Why use a hyperbolic substitution? All you're doing is making the process much longer.

Reply 5

ghostwalker

Original post by ViralRiver

Am I wrong, or are there just multiple ways of doing this?

Differentiate your answer and see what you get.

Reply 6

Farhan.Hanif93

Original post by olipal

Hmm, I guess for a question like this I would probably play it safe and just complete the square in the denominator. Unless they specifically ask for a substitution, that's normally the best move.

Original post by mir3a

Complete the square, and let u=x+2. Why use a hyperbolic substitution? All you're doing is making the process much longer.

That's assuming he has a formula book to hand or that he remembers the standard integrals. Which may or may not be the case.

Original post by ViralRiver

Trying to integrate

\frac{1}{\sqrt{x^2+4x+5}}

. My book, and Wolfram give the answer

arsinh(x+2)

, using a hyperbolic substitution, but I get the answer

ln|sec\theta +tan\theta|=ln|x^2+5x+7|

using a trigonometric substitution (

x + 2 = tan\theta

). Am I wrong, or are there just multiple ways of doing this?

Both methods are fine but I think your logarithmic answer is wrong. To check if they're equivalent, note that

\sinh ^{-1}x = \ln (x+\sqrt{x^2+1})

. Therefore the answer

\sinh ^{-1}(x+2) = \ln ((x+2)+\sqrt{x^2+4x+5})

, which is not what you got for the logarithm version.

Reply 7

ViralRiver

Not sure you're all understanding. The question asked me to use a substitution, hyperbolic or trigonometric. It didn't specify which, but it doesn't allow for a normal algebraic substitution. Farhan, I went through that process myself and realised they weren't the same, but had a feeling it may have been something to do with the constant?

Reply 8

olipal

If that's the case, could it be using a subsitution of say,

x+2=\sinh u

?

Edit: Oh, and for the record the substitution you have chosen to use works too, but as farhan says you haven't expressed

\sec \theta

in terms of

\tan \theta

correctly. Once you have done this the logarithmic equivalent of

arsinh (x+2)

will equal your simplified expression.

(edited 12 years ago)

Reply 9

nuodai

How did you get from

\ln |\sec \theta + \tan \theta|

\ln |x^2+5x+7|

? Note that

\sec \theta = \sqrt{1+\tan^2 \theta}

. You should get the same answer as in the book.

To be honest, if you see

\sqrt{1+x^2}

then the substitution

x=\sinh u

is usually safer than

x=\tan \theta

, which is better for just

1+x^2

on its own (or with an integer power).

Reply 10

ViralRiver

I used the substitution

x+2=tan\theta

. This reduced the integrand to

sec\theta

, giving

ln|sec\theta +tan\theta|

. I used

sec\theta =1+tan^2\theta =x^2+4x+5

and substituted it into the logarithm. I can't see anything wrong with it, but I'm sure there's something simple I'm missing, or I integrated incorrectly >< .

Reply 11

nuodai

Original post by ViralRiver

I used the substitution

x+2=tan\theta

. This reduced the integrand to

sec\theta

, giving

ln|sec\theta +tan\theta|

. I used

sec\theta =1+tan^2\theta =x^2+4x+5

and substituted it into the logarithm. I can't see anything wrong with it, but I'm sure there's something simple I'm missing, or I integrated incorrectly >< .