# Integration by substitution

1. Dec 26, 2007

### 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!

2. Dec 26, 2007

### HallsofIvy

Staff Emeritus
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 $sinh^2(u)+ 1= cosh^2(u)$.

3. Dec 26, 2007

### arildno

I would have, indeed..

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

Last edited: Dec 26, 2007
4. Dec 26, 2007

### unplebeian

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

That should help.

5. Dec 26, 2007

### 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.

6. Dec 26, 2007

### 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?

7. Dec 26, 2007

### unplebeian

First of all, post this as a seperate 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.

Last edited: Dec 26, 2007
8. Dec 27, 2007

### Gib Z

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.

9. Dec 28, 2007

### rohanprabhu

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

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10. Dec 29, 2007

### 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.

11. Dec 29, 2007

### rohanprabhu

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

12. Dec 29, 2007

### 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.

Last edited: Dec 29, 2007
13. Dec 29, 2007

### rohanprabhu

i hate myself.. and u :p

14. Dec 29, 2007

### Mute

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

15. Dec 29, 2007