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\int_0^π/2 {cos x dx/(1+sin^2x)} :/
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Let sin^2 x = y and then solve
D/DX SIN^2X=?
y = sin x would be a better choice
@hartnn yeah i agree
Isn't there a 2 coefficient?
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oh that is pi/2 upper limit, never mind
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
u get it??
;)
the limits would change
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thak u !!!
hint it's tan^-1 x (antidreivativee) and yeah I messed up the limits will change
u = sin x when x = 0, u= 0 when x= pi/2, u= 1 new limits for u 0 to 1
@hartnn ty
give me a metal plz
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ty :)
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