# Integration by substitution

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
Homework Helper
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
Homework Helper
Gold Member
Dearly Missed
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:
$$I=Tanh(u)=Tanh(Sinh^{-1}(x))=\frac{x}{\sqrt{1+x^{2}}}$$

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

That should help.

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.

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

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 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.

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Gib Z
Homework Helper
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.

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

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..}

#### Attachments

• integral_q1.jpg
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Mute
Homework Helper
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.

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
Homework Helper
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.

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

i hate myself.. and u :p

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

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

Gib Z
Homework Helper
i hate myself.. and u :p

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