Solving an Integral Involving Conversion to Cosine

[mugzieee]

i had this problem on an exam today,
Intagral of (x^2)/(4+x^2)^(7/2)
i arrived at a point where i had to convert the integral of 1/secx to cos, is that correct?

 

[Sirus]

Well, someone here would have to do the whole problem to see if that's right. Why don't you show us your work, and we'll go from there.

 

[dextercioby]

Okay.Do first a part integration.Denote the integral u want to compute by "I".

[tex] I=:\int \frac{x^{2}}{(x^{2}+4)^{\frac{7}{2}}} \ dx [/tex] (1)

Do a part integration:

[tex] I=(\frac{1}{2}x)[\frac{(x^{2}+4)^{-\frac{5}{2}}}{-\frac{5}{2}}]+\frac{1}{5}\int \frac{dx}{(x^{2}+4)^{\frac{5}{2}}} [/tex] (2)

Denote the integral from (2) by J...Make the substitution:

[tex] x=2\sinh t [/tex] (3)

[tex] dx=2\cosh t \ dt [/tex] (4)

Then "J" becomes:

[tex] J=\int \frac{2\cosh t}{4^{\frac{5}{2}} \cosh^{5}t} \ dt=\frac{1}{16} \int \frac{dt}{\cosh^{4}t} [/tex] (5)

Now make the substitutions:

[tex] \tanh t= u [/tex] (6)

[tex] dt=\frac{du}{1-u^{2}} [/tex] (7)

So "J" will finally be

[tex] J=\frac{1}{16}\int (1-u^{2}) \ du =\frac{1}{16}(u-\frac{u^{3}}{3} +C) [/tex] (8)

Now express "J" in terms of "x" & plug in (2)...

Daniel.

 

[dextercioby]

Okay,here are a part of the missing calculations,in case you could't do it,even after having gotten the indications...

Daniel.

 

[Jameson]

If you don't want to use hyperbolic susbtitutions, you could use the tabular method. Choose [tex]x^2[/tex] as the term you differentiate and use [tex]\frac{1}{({x^2+4})^\frac{7}{2}}[/tex] as the term you integrate.

 

[dextercioby]

What's the tabular method (i've never heard of this name) ?

Daniel.

 

[Data]

The table method is an algorithm for doing multiple integrations by parts quickly. It's often emphasized to engineers over simply doing the integrations in the normal way (don't ask me why, because it's not any easier when you're doing the integration by hand anyways, and it makes you forget why the method works at all).

 

[dextercioby]

No wonder i haven't heard of it.I'm not (and never will be) an engineer...

Daniel.

 

[Data]

Here's an alternate way of doing the integral, by the way, without resorting to hyperbolic functions and integration by parts, and that involves using [tex]{1\over \sec{x}} = \cos{x}[/tex]:


[tex] I(x) = \int \frac{x^2}{(x^2+4)^{\frac{7}{2}}} dx = {1\over 16}\int \frac{\tan^2{t}\sec^2{t}}{\sec^7{t}} dt = {1\over 16}\int \frac{\sec^2{t}-1}{\sec^5{t}} dt = {1\over 16}\int (\cos^3{t} - \cos^5{t}) dt = {1\over 16}\int (1-u^2-(1-u^2)^2) du[/tex]

[tex]= {1\over 16} \int (u^2 - u^4) du = {1\over 16}\left( \frac{u^3}{3} - \frac{u^5}{5} + C\right) = {1\over 16}\left( \frac{x^3}{3(4+x^2)^{\frac{3}{2}}} - \frac{x^5}{5(4+x^2)^{\frac{5}{2}}} + C\right)[/tex]


where [tex]x = 2\tan{t}, \; u = \sin{t}[/tex].

 

[dextercioby]

Very good.Now the OP has 2 options.One more advice,learn to break the lines,instead of one big code between [tex] tags,use less code inside 2 tags.It will look better.

Daniel.

P.S.I've always liked hyperbolic functions...

 

[Data]

I like them too. The last time I tried to use them, though, my Mathematical Methods TA decided to take off marks because she didn't want to multiply out the exponentials to see if my answer (expressed in terms of hyperbolic functions) matched hers (even though I provided proof that the transformation I used was right, since it wasn't taught in class - perplexing). I got the marks back of course, but now I'm scared of those things! :)

 

[physicsCU]

If I may interject about the tabular method.

I am an aerospace engineering major, my math classes are from the Applied Math department, same college.

We never learned tabular, my book barely mentions it too.

 

[Data]

All I know is that they teach it here to engineers, and most of them swear by it. I don't understand why~

 

[Sirus]

Perhaps you guys should have waited for the OP's response rather than doing the problem for him/her.