Thread: 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!!

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: , then:

.

So the integral becomes:

.

This can now be integrated by parts to give:



Now we need the standard integral:



which when substituted into gives:

.

To complete this you need to know that:



RonL

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:



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!

Integration by parts is often expressed as . We could also express the LHS as where the indefinite integral of g is the function G to write . If we write then we have the formulation .

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:



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