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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 = .... ?

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

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?

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

and where do i put te integral sign ?

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 :)