anonymous
  • anonymous
definite integral from -pi/2 to 0 (2+tan1/2)sec^2 t/2 dt
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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RadEn
  • RadEn
there tan1/2 t right ?
anonymous
  • anonymous
sorry, it is actually tan t/2
RadEn
  • RadEn
ok, use by integral substitution u = tan 1/2 t du = .... ?

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

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