# Finding the integral

• Jan 1st 2012, 05:57 PM
Furyan
Finding the integral
Hello

Just when I thought I was making some progress. I find that I have no idea at all how to approach finding this integral:

$\displaystyle \int\dfrac{\sec^2{x}}{(1 + \tan{x})^3}$

A nudge in the right direction would be very much appreciated.

Thank you.

P.S Is this site compromised? Firefox says that it is and I had serious problems trying to post the other day. I recommended the site to a friend of mine but she said she will not use it because Chrome says it's a security risk. That's such a shame, there are so many brilliant people here who can help people who are stuggling with maths. I tried posting this in the appropriate foum, but was not allowed to do so.
• Jan 1st 2012, 05:59 PM
Prove It
Re: Finding the integral
Quote:

Originally Posted by Furyan
Hello

Just when I thought I was making some progress. I find that I have no idea at all how to approach finding this integral:

$\displaystyle \int\dfrac{\sec^2{x}}{(1 + \tan{x})^3}$

A nudge in the right direction would be very much appreciated.

Thank you.

P.S Is this site compromised? Firefox says that it is and I had serious problems trying to post the other day. I recommended the site to a friend of mine but she said she will not use it because Chrome says it's a security risk. That's such a shame, there are so many brilliant people here who can help people who are stuggling with maths. I tried posting this in the appropriate foum, but was not allowed to do so.

Make the substitution \displaystyle \displaystyle \begin{align*} u = 1 + \tan{x} \implies du = \sec^2{x}\,dx \end{align*}.
• Jan 2nd 2012, 01:19 PM
Furyan
Re: Finding the integral
Thank you Prove It

Quote:

Originally Posted by Prove It
Make the substitution \displaystyle \displaystyle \begin{align*} u = 1 + \tan{x} \implies du = \sec^2{x}\,dx \end{align*}.

I have the following:

$\displaystyle \int\sec^2x(1 + tanx)^{-3} dx$

$\displaystyle u = 1 + tanx$

$\displaystyle \int\sec^2xu^{-3} dx =\int\sec^2xu^{-3}\dfrac{dx}{du} du$

$\displaystyle \dfrac{du}{dx} = \sec^2x$

$\displaystyle \dfrac{dx}{du} = \dfrac{1}{\sec^2x}$

$\displaystyle \int\sec^2xu^{-3}\dfrac{1}{\sec^2x} du = \int u^{-3}du$

= $\displaystyle -\dfrac{1}{2}u^{-2} +c$

$\displaystyle u = 1 + tanx$

$\displaystyle \int\sec^2x(1 + tanx)^{-3} dx = -\dfrac{1}{2}(1+tanx)^{-2} + c$

I stumbled through that a bit.

Thank you for getting me started.
• Jan 2nd 2012, 04:40 PM
Prove It
Re: Finding the integral
Your answer is correct, though you took the scenic route. You can see that \displaystyle \displaystyle \begin{align*} \sec^2{x}\,dx \end{align*} is already in the integral. Replace it with \displaystyle \displaystyle \begin{align*} du \end{align*}.
• Jan 2nd 2012, 05:56 PM
Furyan
Re: Finding the integral
Quote:

Originally Posted by Prove It
Your answer is correct, though you took the scenic route. You can see that \displaystyle \displaystyle \begin{align*} \sec^2{x}\,dx \end{align*} is already in the integral. Replace it with \displaystyle \displaystyle \begin{align*} du \end{align*}.

Thank you, I see what you are saying now, but could someone please clarify this:

I understood that $\displaystyle \dfrac{du}{dx}$ does not mean $\displaystyle du$ divided by $\displaystyle dx$ so why can I rearrange$\displaystyle \dfrac{du}{dx} = \sec^2x$ to get $\displaystyle du = \sec^2x dx$.

And if $\displaystyle \dfrac{d}{dx}$ is an operator and $\displaystyle \dfrac{du}{dx}$ means the derivative of $\displaystyle u$ with respect to $\displaystyle x$. What does $\displaystyle du = \sec^2x dx$ mean?

Thank you.
• Jan 2nd 2012, 06:03 PM
Prove It
Re: Finding the integral
Quote:

Originally Posted by Furyan
Thank you I get that now, but could someone clarify this:

I understood that $\displaystyle \dfrac{du}{dx}$ does not mean $\displaystyle du$ divided by $\displaystyle dx$ so why can I rearrange$\displaystyle \dfrac{du}{dx} = \sec^2x$ to get $\displaystyle du = \sec^2x dx$.

And if $\displaystyle \dfrac{d}{dx}$ is an operator and $\displaystyle \dfrac{du}{dx}$ means the derivative of $\displaystyle u$ with respect to $\displaystyle x$. What does $\displaystyle du = \sec^2x dx$ mean.

It's a sloppy shorthand, I'll give you that.

Really, what happens is that when you replace \displaystyle \displaystyle \begin{align*} \sec^2{x} \end{align*} with \displaystyle \displaystyle \begin{align*} \frac{du}{dx} \end{align*}, in your integral you get

\displaystyle \displaystyle \begin{align*} \int{\frac{\sec^2{x}}{(1 + \tan{x})^3}\,dx} &= \int{\frac{1}{u^3}\,\frac{du}{dx}\,dx} \\ &= \int{\frac{1}{u^3}\,du} \end{align*}

which simplifies from the Chain Rule (not from "cancelling the \displaystyle \displaystyle \begin{align*} dx \end{align*}"). However, most people do the sloppy simplification to save time.
• Jan 2nd 2012, 06:18 PM
Furyan
Re: Finding the integral
Quote:

Originally Posted by Prove It
It's a sloppy shorthand, I'll give you that.

Really, what happens is that when you replace \displaystyle \displaystyle \begin{align*} \sec^2{x} \end{align*} with \displaystyle \displaystyle \begin{align*} \frac{du}{dx} \end{align*}, in your integral you get

\displaystyle \displaystyle \begin{align*} \int{\frac{\sec^2{x}}{(1 + \tan{x})^3}\,dx} &= \int{\frac{1}{u^3}\,\frac{du}{dx}\,dx} \\ &= \int{\frac{1}{u^3}\,du} \end{align*}

which simplifies from the Chain Rule (not from "cancelling the \displaystyle \displaystyle \begin{align*} dx \end{align*}"). However, most people do the sloppy simplification to save time.

Thank you very much :)