## anonymous 3 years ago definite integral from -pi/2 to 0 (2+tan1/2)sec^2 t/2 dt

there tan1/2 t right ?

2. anonymous

sorry, it is actually tan t/2

ok, use by integral substitution u = tan 1/2 t du = .... ?

opss.. u = 2+tan1/2 t, i meant :)

5. anonymous

where did you take the half from?

u = 2 + tan1/2 t du = 0 + 1/2 sec^2 1/2 t dt right ?

7. anonymous

definite integral from -pi/2 to 0 (2+tan t/2)sec^2 t/2 dt

8. Kainui

When deciding what to make your substitution with, you should consider what in the integral has its derivative in there too. This sort of is the process of reversing the chain rule. In this case you should remember that the derivative of tangent is secant squared.

agreed ^

10. anonymous

so where did u get that 1/2 from? I did not get it, sorry

11. Kainui

$\frac{ t }{ 2 }=\frac{ 1 }{ 2 }t$

12. Kainui

In the same way 4/2 is equal to 4*(1/2)

13. anonymous

oh okay,

by using the chain rule : if y=tanAx then y'=A*sec^2 Ax

with A is a constant

16. anonymous

okay

so, the integration will be int 2u du right ?

18. anonymous

ye

ok, what's the result it ...?

int 2u du = ... ?

21. anonymous

sec^2 i/2 t^2/2

int (2u du) = u^2 right ?

23. anonymous

I am confused,

24. anonymous

i got it,

25. anonymous

|dw:1355891028471:dw|

26. anonymous

yes, and then?

27. Kainui

Assuming this is right, you need to remember that the problem you were given to solve had no "u" in it at all, so you must plug in "u" back to the equation to get it in terms of "x".

28. anonymous

so is that ( tan 1/2 t)^2 ?

(2 + tan 1/2 t)^2 now, plug and calculate for t = 0 and t =-pi/2

30. anonymous

tan of 0= 0 and tan of -pi/4 = -1

31. anonymous

(2 + tan 1/2 (0))^2 - (2 + tan 1/2 (-pi/2))^2 = (2+tan(0))^2 - (2+tan(-pi/4)^2 = (2+0)^2 - (2 - tan(pi/4))^2 = 4 - (2 - 1)^2 = 4 - 1 = 3 if i dont make mistake :)

32. anonymous

and where do i put te integral sign ?

33. anonymous

just remember that int (f(x)) dx [a,b] = [F(x)] [a,b] = F(b)-F(a) after u integrate it, u got (2 + tan 1/2 t)^2 right ? now, substitute for highest interval (t=0) and then subtract with for t=-pi/4 (low interval)

34. anonymous

okay,

35. anonymous

do u have the options ?

36. anonymous

options?

37. anonymous

multiple choices ?

38. anonymous

no, but i used the wolfram and shows the ans is 5

39. anonymous

i have another question, how would I put this problem in wolframaalpha? |dw:1355919813662:dw|

40. anonymous

sorry, i cant do it in wolfram :)