AP question: Integral of 9 / (1 +9t^2)

[grapez]

A purported AP questions asks to evaluate the following integral:

Integral of (9)/ (1 +9t^2) as 0 < t < 1/3. I understand that we need to use some sort of substitution techniques to solve. But I was hoping to get a specific explanation of what technique to use? Thanks.

 

[MarkFL]

Hello and welcome to MHB, grapez! :D

We are given to evaluate:

\(\displaystyle I=\int_0^{\frac{1}{3}}\frac{9}{1+9t^2}\,dt\)

Any time you see the sum of two squares in the denominator, and a constant in the numerator, think of using the tangent function. What we want to do is let:

\(\displaystyle \tan(\theta)=3t\,\therefore\,\sec^2(\theta)\,d\theta=3\,dt\)

And our integral becomes (we then use $\theta=\arctan(3t)$ to change the limits according to our substitution):

\(\displaystyle I=3\int_0^{\frac{\pi}{4}}\frac{\sec^2(\theta)}{1+\tan^2(\theta)}\,d\theta\)

Observing that we may use the Pythagorean identity \(\displaystyle 1+\tan^2(\theta)=\sec^2(\theta)\) to simplify our integrand, we obtain:

\(\displaystyle I=3\int_0^{\frac{\pi}{4}}\,d\theta=\frac{3\pi}{4}\)

 

[grapez]

Hello and welcome to MHB, grapez! :D

We are given to evaluate:

\(\displaystyle I=\int_0^{\frac{1}{3}}\frac{9}{1+9t^2}\,dt\)

Any time you see the sum of two squares in the denominator, and a constant in the numerator, think of using the tangent function. What we want to do is let:

\(\displaystyle \tan(\theta)=3t\,\therefore\,\sec^2(\theta)\,d\theta=3\,dt\)

And our integral becomes (we then use $\theta=\arctan(3t)$ to change the limits according to our substitution):

\(\displaystyle I=3\int_0^{\frac{\pi}{4}}\frac{\sec^2(\theta)}{1+\tan^2(\theta)}\,d\theta\)

Observing that we may use the Pythagorean identity \(\displaystyle 1+\tan^2(\theta)=\sec^2(\theta)\) to simplify our integrand, we obtain:

\(\displaystyle I=3\int_0^{\frac{\pi}{4}}\,d\theta=\frac{3\pi}{4}\)

Thanks for all the help. A couple of questions:

a. How did you factor out the 9 in the denominator? I see you brought a 3 out of the integral, but it is not clear to me how that cancels out the 9?

b. A more general question about writing in the forum - how does one code the various operations (integrands, fractions, etc)? The forum has a unique coding language that allows one to type such functions?

 

[MarkFL]

a. How did you factor out the 9 in the denominator? I see you brought a 3 out of the integral, but it is not clear to me how that cancels out the 9?

The original integral may be written as:

\(\displaystyle I=3\int_0^{\frac{1}{3}}\frac{3}{1+(3t)^2}\,dt\)

Now it is more clear how the substitution I suggested works. :)

b. A more general question about writing in the forum - how does one code the various operations (integrands, fractions, etc)? The forum has a unique coding language that allows one to type such functions?

We use $\LaTeX$, which is a standard markup language for mathematics. We have a tutorial here on how to get started with it, and there are many tutorials online.

We have some custom tools available here to aid you in the construction of such markup. To the right of the editor, you will find a Quick $\LaTeX$ tool which you can use to generate specific symbols and commands, and then below the editor you will find our $\LaTeX$ Live Preview that will allow you to quickly preview your expressions before copying them to your post.

 

[grapez]

The original integral may be written as:

\(\displaystyle I=3\int_0^{\frac{1}{3}}\frac{3}{1+(3t)^2}\,dt\)

Now it is more clear how the substitution I suggested works. :)
We use $\LaTeX$, which is a standard markup language for mathematics. We have a tutorial here on how to get started with it, and there are many tutorials online.

We have some custom tools available here to aid you in the construction of such markup. To the right of the editor, you will find a Quick $\LaTeX$ tool which you can use to generate specific symbols and commands, and then below the editor you will find our $\LaTeX$ Live Preview that will allow you to quickly preview your expressions before copying them to your post.

Great, I think I see it now. So is this a common pattern/question that comes up (a constant as a numerator and the sum of two squares in the denominator)? I am wondering if this is something that comes up often in an AP calculus test, where I should notice the trig identity 1 + tan^2(x) = sec^2(x) to cancel out in the integral?

Also, regarding Latex, is this a coding tool that is often often in academia? Is it worthwhile that I invest time in learning the tool to type out texts in my future studies?

 

[MarkFL]

Great, I think I see it now. So is this a common pattern/question that comes up (a constant as a numerator and the sum of two squares in the denominator)? I am wondering if this is something that comes up often in an AP calculus test, where I should notice the trig identity 1 + tan^2(x) = sec^2(x) to cancel out in the integral?

Let's look at the general indefinite integral:

\(\displaystyle \int\frac{a}{b+cx^2}\,dx\)

We can rewrite this as:

\(\displaystyle \frac{a}{b}\int\frac{1}{1+\left(\sqrt{\dfrac{c}{b}}x\right)^2}\,dx\)

Now we have the sum of two squares in the denominator of the integrand, and the numerator is a constant.

So, we see we should use the substitution:

\(\displaystyle \sqrt{\frac{c}{b}}x=\tan(\theta)\,\therefore\,dx=\sqrt{\frac{b}{c}}\sec^2(\theta)\,d\theta\)

And so now we have (after making the substitution and applying the Pythagorean identity):

\(\displaystyle \frac{a}{\sqrt{bc}}\int\,d\theta=\frac{a}{\sqrt{bc}}\theta+C=\frac{a}{\sqrt{bc}}\arctan\left(\sqrt{\frac{c}{b}}x\right)+C\)

Also, regarding Latex, is this a coding tool that is often often in academia? Is it worthwhile that I invest time in learning the tool to type out texts in my future studies?

Yes, $\LaTeX$ is very useful to know, and is a great way to create documents with pretty mathematics. :)