# Integration Help

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• May 5th 2006, 08:23 PM
AfterShock
Integration Help
Hello,

I was wondering if someone could help me solve the integral of the following:

int(sec^2(sqrt(x))dx).

Now, I tried letting u = sqrt(x), but that gets me nowhere; I wouldn't be able to integrate by parts, as what would I let dv be? Trying to let u = the whole function does little good...maybe I am missing something (well, obviously I am!).

Ugh.

Thanks in advance for your help!!
• May 5th 2006, 09:43 PM
CaptainBlack
Quote:

Originally Posted by AfterShock
Hello,

I was wondering if someone could help me solve the integral of the following:

int(sec^2(sqrt(x))dx).

Now, I tried letting u = sqrt(x), but that gets me nowhere; I wouldn't be able to integrate by parts, as what would I let dv be? Trying to let u = the whole function does little good...maybe I am missing something (well, obviously I am!).

Ugh.

Thanks in advance for your help!!

First change the variable, let: $u=\sqrt{x}$, then:

$
dx=2u\ du
$
.

So the integral becomes:

$
\int (\sec( \sqrt{x})^2 dx=\int \frac{2u}{(\cos(u))^2}du
$
.

This can now be integrated by parts to give:

$
\int \frac{2u}{(\cos(u))^2}du=$
$2 u\int \frac{1}{(\cos(u))^2}du-2\int \left\{\int}\frac{1}{(\cos(u))^2}du\right\}\ du\ \ \ \dots(1)
$

Now we need the standard integral:

$
\int \frac{1}{(\cos(u))^2}du=\tan(u)
$

which when substituted into $(1)$ gives:

$
\int \frac{2u}{(\cos(u))^2}du=$
$2 u\tan(u)-2\int \tan(u) du
$
.

To complete this you need to know that:

$
\int \tan(x) dx = -\ln(\cos(x))
$

RonL
• May 5th 2006, 11:18 PM
AfterShock
Thank you CaptainBlack. I have a few questions regarding it, though, since I truly want to know how to do it.

I get the first part, but I lose you on this part:

This can now be integrated by parts to give:

$
2 u\int \frac{1}{(\cos(u))^2}du-2\int \left\{\int}\frac{1}{(\cos(u))^2}du\right\}\ du\ \ \ \dots(1)
$

I didn't know you could just pull the 2u out like that; and then you have an integral of an integral? I thought you were doing integration by parts (uv - int(vdu)). I believe I understand everything after that line.

Thanks- it's appreciated!
• May 5th 2006, 11:56 PM
rgep
Integration by parts is often expressed as $\int u dv = [uv] - \int v du$. We could also express the LHS as $\int fg dx$ where the indefinite integral of g is the function G to write $\int fg = fG - \int f'G$. If we write $G = \int g$ then we have the formulation $\int fg = f \int G - \int f' \int G$.
• May 6th 2006, 12:00 AM
CaptainBlack
Quote:

Originally Posted by AfterShock
Thank you CaptainBlack. I have a few questions regarding it, though, since I truly want to know how to do it.

I get the first part, but I lose you on this part:

This can now be integrated by parts to give:

$
2 u\int \frac{1}{(\cos(u))^2}du-2\int \left\{\int}\frac{1}{(\cos(u))^2}du\right\}\ du\ \ \ \dots(1)
$

I didn't know you could just pull the 2u out like that; and then you have an integral of an integral? I thought you were doing integration by parts (uv - int(vdu)). I believe I understand everything after that line.

Thanks- it's appreciated!

In your notation int(u dv)=uv - int(vdu), though as we are using u as a
variable let us write this as in(f dg)=f g - int(g df), so here we are putting
f(u)=2u, and dg(u)=1/cos^2(u), then g(u)=int(1/cos^2(u)), and df(u)=2.

Then the result follows.

RonL