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

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1. RadEn

there tan1/2 t right ?

2. roselin

sorry, it is actually tan t/2

3. RadEn

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

4. RadEn

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

5. roselin

where did you take the half from?

6. RadEn

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

7. roselin

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.

9. RadEn

agreed ^

10. roselin

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. roselin

oh okay,

14. RadEn

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

15. RadEn

with A is a constant

16. roselin

okay

17. RadEn

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

18. roselin

ye

19. RadEn

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

20. RadEn

int 2u du = ... ?

21. roselin

sec^2 i/2 t^2/2

22. RadEn

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

23. roselin

I am confused,

24. roselin

i got it,

25. roselin

|dw:1355891028471:dw|

26. roselin

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. roselin

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

29. RadEn

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

30. roselin

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

31. tanjung

(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

and where do i put te integral sign ?

33. tanjung

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

okay,

35. tanjung

do u have the options ?

36. roselin

options?

37. tanjung

multiple choices ?

38. roselin

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

39. roselin

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

40. tanjung

sorry, i cant do it in wolfram :)

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