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roselinBest ResponseYou've already chosen the best response.0
sorry, it is actually tan t/2
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
ok, use by integral substitution u = tan 1/2 t du = .... ?
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
opss.. u = 2+tan1/2 t, i meant :)
 one year ago

roselinBest ResponseYou've already chosen the best response.0
where did you take the half from?
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
u = 2 + tan1/2 t du = 0 + 1/2 sec^2 1/2 t dt right ?
 one year ago

roselinBest ResponseYou've already chosen the best response.0
definite integral from pi/2 to 0 (2+tan t/2)sec^2 t/2 dt
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
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.
 one year ago

roselinBest ResponseYou've already chosen the best response.0
so where did u get that 1/2 from? I did not get it, sorry
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
\[\frac{ t }{ 2 }=\frac{ 1 }{ 2 }t\]
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
In the same way 4/2 is equal to 4*(1/2)
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
by using the chain rule : if y=tanAx then y'=A*sec^2 Ax
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
so, the integration will be int 2u du right ?
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
ok, what's the result it ...?
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
int (2u du) = u^2 right ?
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
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".
 one year ago

roselinBest ResponseYou've already chosen the best response.0
so is that ( tan 1/2 t)^2 ?
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
(2 + tan 1/2 t)^2 now, plug and calculate for t = 0 and t =pi/2
 one year ago

roselinBest ResponseYou've already chosen the best response.0
tan of 0= 0 and tan of pi/4 = 1
 one year ago

tanjungBest ResponseYou've already chosen the best response.0
(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 :)
 one year ago

roselinBest ResponseYou've already chosen the best response.0
and where do i put te integral sign ?
 one year ago

tanjungBest ResponseYou've already chosen the best response.0
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)
 one year ago

tanjungBest ResponseYou've already chosen the best response.0
do u have the options ?
 one year ago

roselinBest ResponseYou've already chosen the best response.0
no, but i used the wolfram and shows the ans is 5
 one year ago

roselinBest ResponseYou've already chosen the best response.0
i have another question, how would I put this problem in wolframaalpha? dw:1355919813662:dw
 one year ago

tanjungBest ResponseYou've already chosen the best response.0
sorry, i cant do it in wolfram :)
 one year ago
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