## anonymous one year ago Integration help

1. anonymous

|dw:1442609775803:dw|

2. anonymous

@Empty

3. anonymous

i have the answer as 9.747 rounded to 3 decimal places but i dont understand how to get

4. Zarkon

$x=\frac{\tan(\theta)}{2}$

5. anonymous

????

6. anonymous

7. zepdrix

It's hard to take the root of a sum... that addition sign is causing a problem. We would like to apply some type of trigonometric substitution because our trig identities allow us to get rid of addition/subtraction. $$\rm 1-\sin^2x=\cos^2x$$ $$\rm 1+\tan^2x=\sec^2x$$ And the others..

8. zepdrix

Our problem involves something of the form: $$\rm 1+stuff^2$$ Which is telling us that we want to use the tangent identity, ya?

9. zepdrix

$\large\rm \int\limits\sqrt{1+4x^2}~dx=\int\limits\sqrt{1+(2x)^2}~dx$We would like to turn the 2x into tangent, then we would have 1+tangent^2 under the root, and would be able to apply our identity!

10. zepdrix

So, as Zarkon indicated, make the substitution, $$\large\rm 2x=\tan\theta$$

11. zepdrix

So we're making the substitution: $$\large\rm \color{orangered}{2x=\tan\theta}$$ Then our integral becomes:$\large\rm \int\limits\limits\sqrt{1+(\color{orangered}{2x})^2}~dx=\int\limits\limits\sqrt{1+(\color{orangered}{\tan\theta})^2}~dx$

12. zepdrix

We need to also replace the dx with something involving d(theta) though.

13. zepdrix

Whatchu think mage? :O Too confusing?