# Integration problem

## Homework Statement

Determine $$\int$$4dy/(1+9y$$^{2}$$) With limits of 2,0.

## The Attempt at a Solution

Have attempted ingtegration by substitution but have had no luck solving this problem. A maths tutor who went over it very quickly established there was a tan in the answer, i have not integrated anything like this before so don't really know where to start.

## The Attempt at a Solution

do you know what the derivative of arctan is?

maybe this looks a little more familiar

$$\int_{0}^{2}\frac{4dy}{1+(3y)^{2}}$$

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I have just looked up the definition, can't quite see how it will fit

$$4\int_{0}^{2}\frac{dy}{1+(3y)^{2}}$$

$$\mbox{Let u=3y}$$

does it look a little more familiar now?

I ended up with

4arctan(6)

Am i close?

no, example

$$\frac{d}{dy}\tan^{-1}(3y^{2})=\frac{6ydy}{1+(3y^{2})^{2}}$$

Hmm i can't seem to get it, when i integrate i get

$$\frac{1}{12}tan^{-1}(12)$$

Ignore that last post, is

$$\frac{4}{3}tan^{-1}(6)$$ correct?

do you notice the pattern with my problem?

the angle is $$3y^{2}$$

where did my angle and derivative end up when i differentiated?

Ignore that last post, is

$$\frac{4}{3}tan^{-1}(6)$$ correct?
you're constants are correct but you're angle is wrong. if i took the derivative of your problem it would end up being 0 b/c you're basically saying it's a constant.

$$\frac{4}{3}\frac{0}{1+36}$$

The 6 is just the value of the limits substituted in to get a final answer, or is not that what the substituted value would be?

The 6 is just the value of the limits substituted in to get a final answer, or is not that what the substituted value would be?
yes that is correct, i did not realize you were already plugging your limits in and evaluating. sorry, miscommunication.

no problem, thank you very much for your assistance :)