# Thread: Integration Help

1. ## 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!!

2. 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: $\displaystyle u=\sqrt{x}$, then:

$\displaystyle dx=2u\ du$.

So the integral becomes:

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

This can now be integrated by parts to give:

$\displaystyle \int \frac{2u}{(\cos(u))^2}du=$$\displaystyle 2 u\int \frac{1}{(\cos(u))^2}du-2\int \left\{\int}\frac{1}{(\cos(u))^2}du\right\}\ du\ \ \ \dots(1) Now we need the standard integral: \displaystyle \int \frac{1}{(\cos(u))^2}du=\tan(u) which when substituted into \displaystyle (1) gives: \displaystyle \int \frac{2u}{(\cos(u))^2}du=$$\displaystyle 2 u\tan(u)-2\int \tan(u) du$.

To complete this you need to know that:

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

RonL

3. 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:

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

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!

4. Integration by parts is often expressed as $\displaystyle \int u dv = [uv] - \int v du$. We could also express the LHS as $\displaystyle \int fg dx$ where the indefinite integral of g is the function G to write $\displaystyle \int fg = fG - \int f'G$. If we write $\displaystyle G = \int g$ then we have the formulation $\displaystyle \int fg = f \int G - \int f' \int G$.

5. 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:

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

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