## anonymous 5 years ago how do i put ln[ sectheta + tantheta] in terms of x

1. anonymous

a. There's no "x" in this expression. b. In order to put it in terms of a variable; it has to be in an equation form.

2. anonymous

the ogrinal equation from solving it was intergal of dx/1-x^2 if that helps

3. anonymous

true, isn't there a given x np?

4. anonymous

HaHaHA (breaking down due to extreme lough)

5. anonymous

lol, what's so funny angoo?

6. anonymous

I don't know

7. anonymous

Ahhh, so you used a trig substitution. Now, go back to whatever value you used for x to solve the integral (in terms of theta), solve for sec(theta) and tan(theta) by drawing out the triangles, and put those values - in terms of x - in your final expression.

8. anonymous

>_> seriously AG

9. anonymous

just wrote that to make you smile

10. anonymous

did you understand what Quantum said np65? ^_^

11. anonymous

and it made me smile :)

12. anonymous

so i have the triangle would that put me at 1-x^2+x then?

13. anonymous

14. anonymous

what was your value of x when you integrated?

15. anonymous

What trig substitution for x did you use to solve the integral?

16. anonymous

i set x=sin

17. anonymous

how did you set x = sin?

18. anonymous

alright i have the intergal of dx1-x^2 i set u =sintheta and du = costheta dtheta i then subed in u and du

19. anonymous

following that i got dtheta/costheta

20. anonymous

oh I got it!$x = \sin \theta$ and you know that sin x = opposite/hyp your opposite in this case = x hyp = 1 draw the triangle and you'll be able to continue from here ^_^

21. anonymous

hence :$\sec \theta= 1/\cos \theta$ $\tan \theta = \sin \theta/ \cos \theta$ you have sin (theta), now find cos (theta) :) is it clear now?

22. anonymous

so then the other side would be 1-x^2 or use the p-thag therom to get the last side making it the sqrroot(-x^2+1)

23. anonymous

use the pyth.theorem :) that's it. you just don't have x, find it by applying the theorem , then find cos (theta)

24. anonymous

so the last side will be : $\sqrt{1- x^2}$ ^_^

25. anonymous

awesome got it now

26. anonymous

thanks

27. anonymous

np :)