Finding the integral

**Furyan** · January 1st 2012, 05:57 PM

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:

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

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.

**Prove It** · January 1st 2012, 05:59 PM

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:

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

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 $\displaystyle \begin{align*} u = 1 + \tan{x} \implies du = \sec^2{x}\,dx \end{align*}$ .

**Furyan** · January 2nd 2012, 01:19 PM

Thank you Prove It

Originally Posted by Prove It

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

I have the following:

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

$u = 1 + tanx$

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

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

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

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

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

$u = 1 + tanx$

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

I stumbled through that a bit.

Thank you for getting me started.

**Prove It** · January 2nd 2012, 04:40 PM

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

**Furyan** · January 2nd 2012, 05:56 PM

Originally Posted by Prove It

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

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

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

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

Thank you.

**Prove It** · January 2nd 2012, 06:03 PM

Originally Posted by Furyan

Thank you I get that now, but could someone clarify this:

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

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

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

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

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

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

**Furyan** · January 2nd 2012, 06:18 PM

Originally Posted by Prove It

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

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

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

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

Thank you very much

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