# Two substitution integral evaluation questions

• Feb 16th 2009, 12:29 PM
fattydq
Two substitution integral evaluation questions
I've attached a picture, there's a few types of problems I'm having trouble with. I can easily solve simple integral problems involving substitution but these two are stumping me. For the first one I don't even know the deriv. of arc tan or know of a derivative that will be part of something else in the problem.

As for the second one I don't even know what to make u or du because again no part of the problem is the other part's derivative.
• Feb 16th 2009, 12:42 PM
Jameson
1. You should be able to find the derivative of arctan(x) in your calc book or Google it. I would suggest learning to derive it. Either way, it's not a secret to find if you have internet access. Once you look this up, the substitution will be obvious I hope.

2. u=x-3
• Feb 16th 2009, 12:47 PM
Reckoner
For the first, you should know that $\frac d{dx}[\arctan u] = \frac{u'}{1+u^2}$

Do you see this form in your integrand?

For the second one, let $u = x - 3\text.$ Then $du = dx$ and the limits become -3 to 3:

$\int_0^6(x-3)^{23}\,dx = \int_{-3}^3u^{23}\,du$
• Feb 16th 2009, 01:03 PM
fattydq
Quote:

Originally Posted by Jameson
1. You should be able to find the derivative of arctan(x) in your calc book or Google it. I would suggest learning to derive it. Either way, it's not a secret to find if you have internet access. Once you look this up, the substitution will be obvious I hope.

2. u=x-3

1. Ok it's 5/25x^2+5 but that still doesn't divide evenly into 25x^2+1

2.Wow that was easy, stupid oversight.
• Feb 16th 2009, 01:04 PM
fattydq
Quote:

Originally Posted by Reckoner
For the first, you should know that $\frac d{dx}[\arctan u] = \frac{u'}{1+u^2}$

Do you see this form in your integrand?

For the second one, let $u = x - 3\text.$ Then $du = dx$ and the limits become -3 to 3:

$\int_0^6(x-3)^{23}\,dx = \int_{-3}^3u^{23}\,du$

Yes but even knowing that it still doesn't divide evenly into another term in the problem
• Feb 16th 2009, 01:13 PM
Reckoner
Quote:

Originally Posted by fattydq
Yes but even knowing that it still doesn't divide evenly into another term in the problem

Let $u = \arctan5x\Rightarrow du=\frac{5\,dx}{1+(5x)^2}=\frac{5\,dx}{1+25x^2}\te xt.$

Then we have

$\int\frac{\arctan5x}{1+25x^2}dx$

= $\frac15\int u\,du$

Simple, right?

If you wanted, you could still do $u=5x,$ but then you would need to do a second substitution ( $v=\arctan u$).
• Feb 16th 2009, 01:19 PM
fattydq
Quote:

Originally Posted by Reckoner
Let $u = \arctan5x\Rightarrow du=\frac{5\,dx}{1+(5x)^2}=\frac{5\,dx}{1+25x^2}\te xt.$

Then we have

$\int\frac{\arctan5x}{1+25x^2}dx$

= $\frac15\int u\,du$

Simple, right?

If you wanted, you could still do $u=5x,$ but then you would need to do a second substitution ( $v=\arctan u$).

It seems simple to you but I still don't understand. how does 25x^2+1 divide into 25x^2+5 and turn out to be five. I mean 2+4 divided by 2+ 2 isn't 2 just because the 2's can cancel out, if that makes sense.
• Feb 16th 2009, 01:31 PM
Reckoner
Quote:

Originally Posted by fattydq
It seems simple to you but I still don't understand. how does 25x^2+1 divide into 25x^2+5 and turn out to be five. I mean 2+4 divided by 2+ 2 isn't 2 just because the 2's can cancel out, if that makes sense.

Sorry, let me make it more clear.

We are using $u=\arctan5x,$ which gives $du=\frac{5\,dx}{1+25x^2}\Rightarrow dx=\frac15(1+25x^2)\,du\text.$ Directly after the substitution, we have

$\int\frac u{1+25x^2}\left[\frac15(1+25x^2)\,du\right]$

$=\int u\left(\frac15\right)\left(\frac{1+25x^2}{1+25x^2} \right)du\text.$

Now, we can simplify and pull the constant out of the integrand to get

$\frac15\int u\,du\text.$

Does that help? If not, I'm afraid you'll have to wait a few hours before I can respond again, as I'll be leaving in a couple minutes.
• Feb 16th 2009, 01:38 PM
fattydq
I still don't understand where the 1/5 comes from. What do you have to multiply by 1/5? Normally I treat these problems as you want your u's derivative (du) to contain part of the function so that you can divide it out, and 25x^2+1 just doesn't divide nto 25x^5 evenly how does this become 1/5.
• Feb 16th 2009, 01:48 PM
Reckoner
Quote:

Originally Posted by fattydq
I still don't understand where the 1/5 comes from. What do you have to multiply by 1/5? Normally I treat these problems as you want your u's derivative (du) to contain part of the function so that you can divide it out, and 25x^2+1 just doesn't divide nto 25x^5 evenly how does this become 1/5.

Okay, real quick before I go: the $\frac15$ comes from the substitution for $dx\text.$ The reason it is there is because of the use of the chain rule in differentiating $\arctan5x\text.$ The 5 was a result of the chain rule, and when solving for $dx$ it becomes $\frac15,$ which is then inserted into the integral when we substitute.