## roselin Group Title definite integral from -pi/2 to 0 (2+tan1/2)sec^2 t/2 dt one year ago one year ago

there tan1/2 t right ?

2. roselin Group Title

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. roselin Group Title

where did you take the half from?

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

7. roselin Group Title

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

8. Kainui Group Title

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. roselin Group Title

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

11. Kainui Group Title

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

12. Kainui Group Title

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

13. roselin Group Title

oh okay,

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

with A is a constant

16. roselin Group Title

okay

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

18. roselin Group Title

ye

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

int 2u du = ... ?

21. roselin Group Title

sec^2 i/2 t^2/2

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

23. roselin Group Title

I am confused,

24. roselin Group Title

i got it,

25. roselin Group Title

|dw:1355891028471:dw|

26. roselin Group Title

yes, and then?

27. Kainui Group Title

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. roselin Group Title

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. roselin Group Title

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

31. tanjung Group Title

(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. roselin Group Title

and where do i put te integral sign ?

33. tanjung Group Title

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. roselin Group Title

okay,

35. tanjung Group Title

do u have the options ?

36. roselin Group Title

options?

37. tanjung Group Title

multiple choices ?

38. roselin Group Title

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

39. roselin Group Title

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

40. tanjung Group Title

sorry, i cant do it in wolfram :)