## ksaimouli Group Title http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__37.gif one year ago one year ago

1. ksaimouli

@mathmate

2. ksaimouli

i did this with u substituion it did not work

3. ksaimouli

i am getting something like ln2/2

4. mathmate

It looks to me more an atan(2x)

5. ksaimouli

how

6. ksaimouli

i took u=1+4t^2

7. ksaimouli

du/8t=dt

8. mathmate

The standard integral is $$\int \frac{dx}{1+x^2} = atan(x) + C$$

9. ksaimouli

|dw:1358025284718:dw|

10. mathmate

So you'd be integrating $$\int \frac{du}{8u(1+4t^2)}$$

11. ksaimouli

use the equation button

12. mathmate

The result is not ln(u)

13. ksaimouli

ok i should remember that formula right?

14. mathmate

I suggest you use y=2t

15. mathmate

Yes, it's on the first page of standard integrals. Check it out.

16. ksaimouli

ok

17. mathmate

$$\large \int_0^{1/2} \frac{dt}{1+4t^2}= \int_0^{1} \frac{2dy}{1+y^2} = [atan(y)]_0^{1}= 2\pi/4=\pi/2$$

18. ksaimouli

@mathmate i found the formula to be $\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a }+c$

19. mathmate

Right, so a=1/2.

20. ksaimouli

what is x then

21. ksaimouli

$\int\limits_{}^{}\frac{ 1 }{ a^2+x^2}=\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a}+c$

22. mathmate

x is your t. you can replace the identity with $$\large \int \frac{dx}{a^2+x^2} = \int \frac{dt}{a^2(1+(t/a)^2)}$$

23. mathmate

and a=1/2

24. ksaimouli

can i do $\frac{ 1 }{ 1 }\tan^{-1} \frac{ 2t }{ 1}$

25. ksaimouli

and t=1/2

26. mathmate

t is a variable, and you cannot equate it to a constant. Use the formula you found, change t for x, and 1/2 for a, then the left-hand side is exactly your question. All you need to do after that is to evaluate the right-hand side.