# Thread: Two substitution integral evaluation questions

1. ## 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.

2. 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

3. For the first, you should know that $\displaystyle \frac d{dx}[\arctan u] = \frac{u'}{1+u^2}$

Do you see this form in your integrand?

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

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

4. 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.

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

Do you see this form in your integrand?

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

$\displaystyle \int_0^6(x-3)^{23}\,dx = \int_{-3}^3u^{23}\,du$
Yes but even knowing that it still doesn't divide evenly into another term in the problem

6. Originally Posted by fattydq
Yes but even knowing that it still doesn't divide evenly into another term in the problem
Let $\displaystyle u = \arctan5x\Rightarrow du=\frac{5\,dx}{1+(5x)^2}=\frac{5\,dx}{1+25x^2}\te xt.$

Then we have

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

=$\displaystyle \frac15\int u\,du$

Simple, right?

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

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

Then we have

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

=$\displaystyle \frac15\int u\,du$

Simple, right?

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

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.

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

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

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

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

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

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.

9. 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.

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