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

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

Calculus1
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there tan1/2 t right ?
sorry, it is actually tan t/2
ok, use by integral substitution u = tan 1/2 t du = .... ?

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opss.. u = 2+tan1/2 t, i meant :)
where did you take the half from?
u = 2 + tan1/2 t du = 0 + 1/2 sec^2 1/2 t dt right ?
definite integral from -pi/2 to 0 (2+tan t/2)sec^2 t/2 dt
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 ^
so where did u get that 1/2 from? I did not get it, sorry
\[\frac{ t }{ 2 }=\frac{ 1 }{ 2 }t\]
In the same way 4/2 is equal to 4*(1/2)
oh okay,
by using the chain rule : if y=tanAx then y'=A*sec^2 Ax
with A is a constant
okay
so, the integration will be int 2u du right ?
ye
ok, what's the result it ...?
int 2u du = ... ?
sec^2 i/2 t^2/2
int (2u du) = u^2 right ?
I am confused,
i got it,
|dw:1355891028471:dw|
yes, and then?
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".
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
tan of 0= 0 and tan of -pi/4 = -1
(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 :)
and where do i put te integral sign ?
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)
okay,
do u have the options ?
options?
multiple choices ?
no, but i used the wolfram and shows the ans is 5
i have another question, how would I put this problem in wolframaalpha? |dw:1355919813662:dw|
sorry, i cant do it in wolfram :)

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