Integrating (x^2+1)^{-3/2} using Substitution

[uman]

Hi all,

I've been studying calculus out of Tom Apostol's book "Calculus". I'm having troube with the following problem in the section on integration by substitution:

Integrate $$\int(x^2+1)^{-3/2}\,dx$$.

I tried the substitution $$u=x^2+1$$ but it didn't seem to work. I can't see anything else that may help. Any hints or solution would be greatly appreciated!

 

[HallsofIvy]

There are two fairly standard substitutions that can by used for something of the form $x^1+ 1$ inside a square root. Since $sin^2(\theta)+ cos^2(\theta)= 1$, dividing through by $cos^2(\theta)$ gives $sin^2(\theta)/cos^2(\theta)+ 1= 1/cos^2(\theta)$ so that $tan^2(\theta)+ 1= sec^2(\theta)$ which suggests the substitution $x= tan(\theta)$. If arildno have gotten here first, he would have suggested x= sinh(u). That works, and may be simpler, because $cosh^2(u)- sinh^2(u)= 1$ so that $sinh^2(u)+ 1= cosh^2(u)$.

 

[arildno]

There are two fairly standard substitutions that can by used for something of the form $x^1+ 1$ inside a square root. Since $sin^2(\theta)+ cos^2(\theta)= 1$, dividing through by $cos^2(\theta)$ gives $sin^2(\theta)/cos^2(\theta)+ 1= 1/cos^2(\theta)$ so that $tan^2(\theta)+ 1= sec^2(\theta)$ which suggests the substitution $x= tan(\theta)$. If arildno have gotten here first, he would have suggested x= sinh(u). That works, and may be simpler, because $cosh^2(u)- sinh^2(u)= 1$ so that $sinh^2(u)+ 1= cosh^2(u)$.

I would have, indeed..

The result is then, of course:
$$I=Tanh(u)=Tanh(Sinh^{-1}(x))=\frac{x}{\sqrt{1+x^{2}}}$$

 

[unplebeian]

Hint: Use sec^2(x) - tan^2(x)= 1 and then use another subst involving derivative of tan x.

That should help.

 

[uman]

Thanks for the help, all! I used HallsofIvy's suggestion which worked marvelously.

Maybe arildno's would have been simpler, however I've only vaguely heard of the hyperbolic trig functions and don't really know them.

 

[uman]

I'm also having trouble with the following integral: $$\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx$$. Any ideas?

 

[unplebeian]

I'm also having trouble with the following integral: $$\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx$$. Any ideas?

First of all, post this as a separate q. You can use the hint I gave above and then just keep simplifying it, keep simplifying it, and after a while you'll arrive at an expression where it should be clear that you have to make another subst. involving sec(u) + 1

That hint should help.

 

[Gib Z]

I'm also having trouble with the following integral: $$\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx$$. Any ideas?

It works out nicely and exactly analogously with both a circular trig substitution and a hyperbolic trig substitution. Firstly the trig substitution, then a linear one.

 

[rohanprabhu]

$$\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx$$

It works out nicely and exactly analogously with both a circular trig substitution and a hyperbolic trig substitution. Firstly the trig substitution, then a linear one.

I think I have a better way.. [please see attachment] {sorry if my handwriting ain't clear..}

 

[Mute]

You made a mistake doing your final integral in the attachment. The correct result is

$$\int \frac{du}{\sqrt{1+u}} = \sqrt{1 + u} =\sqrt{1 + \sqrt{x^2+1}}$$

(up to a constant of integration, of course).


You can then check by differentiation that this is correct.

 

[rohanprabhu]

You made a mistake doing your final integral in the attachment. The correct result is

$$\int \frac{du}{\sqrt{1+u}} = \sqrt{1 + u} =\sqrt{1 + \sqrt{x^2+1}}$$

(up to a constant of integration, of course).


You can then check by differentiation that this is correct.

oops.. I mistakenly took the power of $1 + u$ to be $\frac{1}{2}$ instead of $\frac{-1}{2}$..

 

[Gib Z]

Circular Trig method

$$\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx$$

Let x= tan u. Then dx = sec^2 u du.

$$\int \frac{\tan u \sec^2 u}{\sqrt{\sec^2 u + \sec^3 u}} du = \int \frac{ (\sec u)'}{\sqrt{1+ \sec u}} du$$.

Now let t= 1+ sec u.
$$\int \frac{1}{ \sqrt{t} } dt = 2\sqrt{t} + C = 2\sqrt{1+ \sec u} + C = 2\sqrt{1+ \sqrt{x^2+1}} + C$$.

------------

Hyperbolic Trig method

$$\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx$$

Let x = sinh u, dx = cosh u du

$$\int \frac{ \sinh u \cosh u }{ \sqrt{\cosh^2 u + \cosh^3 u}} du = \int \frac{(\cosh u)'}{\sqrt{1+ \cosh u}} du$$.

Let t= 1+ cosh u.

$$\int \frac{1}{\sqrt{t}} dt = 2\sqrt{t} + C = 2\sqrt{ 1+ \cosh u} + C = 2\sqrt{1+ \sqrt{1+x^2}} + C$$.PS. You both missed a factor of 2 when evaluating the integral.

 

[rohanprabhu]

PS. You both missed a factor of 2 when evaluating the integral.

i hate myself.. and u :p

 

[Mute]

PS. You both missed a factor of 2 when evaluating the integral.

Eh, factors of 2 between friends aren't important. ; )

 

[Gib Z]

i hate myself.. and u :p

Eh, factors of 2 between friends aren't important. ; )