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Mathematics 18 Online
OpenStudy (anonymous):

\int_0^π/2 {cos x dx/(1+sin^2x)} :/

OpenStudy (atlas):

Let sin^2 x = y and then solve

OpenStudy (anonymous):

D/DX SIN^2X=?

hartnn (hartnn):

y = sin x would be a better choice

OpenStudy (atlas):

@hartnn yeah i agree

OpenStudy (anonymous):

Isn't there a 2 coefficient?

OpenStudy (anonymous):

oh that is pi/2 upper limit, never mind

OpenStudy (anonymous):

d/dx sin^2x = 2(sinx)cosx dx d/dx sinx = cosx dx u = sin x du = cos x dx integral from 0 to 2pi 1 / (1+u^2) du

OpenStudy (anonymous):

u get it??

OpenStudy (anonymous):

;)

hartnn (hartnn):

the limits would change

OpenStudy (anonymous):

thak u !!!

OpenStudy (anonymous):

hint it's tan^-1 x (antidreivativee) and yeah I messed up the limits will change

hartnn (hartnn):

u = sin x when x = 0, u= 0 when x= pi/2, u= 1 new limits for u 0 to 1

OpenStudy (anonymous):

@hartnn ty

OpenStudy (anonymous):

give me a metal plz

OpenStudy (anonymous):

ty :)

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