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Calculus1 6 Online
OpenStudy (anonymous):

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

OpenStudy (raden):

there tan1/2 t right ?

OpenStudy (anonymous):

sorry, it is actually tan t/2

OpenStudy (raden):

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

OpenStudy (raden):

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

OpenStudy (anonymous):

where did you take the half from?

OpenStudy (raden):

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

OpenStudy (anonymous):

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

OpenStudy (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.

OpenStudy (raden):

agreed ^

OpenStudy (anonymous):

so where did u get that 1/2 from? I did not get it, sorry

OpenStudy (kainui):

\[\frac{ t }{ 2 }=\frac{ 1 }{ 2 }t\]

OpenStudy (kainui):

In the same way 4/2 is equal to 4*(1/2)

OpenStudy (anonymous):

oh okay,

OpenStudy (raden):

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

OpenStudy (raden):

with A is a constant

OpenStudy (anonymous):

okay

OpenStudy (raden):

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

OpenStudy (anonymous):

ye

OpenStudy (raden):

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

OpenStudy (raden):

int 2u du = ... ?

OpenStudy (anonymous):

sec^2 i/2 t^2/2

OpenStudy (raden):

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

OpenStudy (anonymous):

I am confused,

OpenStudy (anonymous):

i got it,

OpenStudy (anonymous):

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