Integral of 0 to (pi/2) of : ((sec^2)(x))/((tan(x)-1)^2) dx
Can anyone solve this???

Question

Integral of 0 to (pi/2) of : ((sec^2)(x))/((tan(x)-1)^2) dx
Can anyone solve this???

loser66 · Answer

substitute u = tan (x) -1

anonymous · Answer

What ? Lol

anonymous · Answer

Oh. Sorry :0

loser66 · Answer

not too late, fix it, friend

anonymous · Answer

I Did

anonymous · Answer

hey Loser, here's what I get:  
integral 1/u^2du
= -1/u + C
= 1/1-tan(x) + C
= sin(x)/cos(x)-sin(x) evaluated at 0 and pi/2
= -1 - 0 = -1
Does this make any sense?

loser66 · Answer

we don't have C, right?

anonymous · Answer

right, Im just first evaluating as an indefinite integral

loser66 · Answer

I don't know why you have to convert to sin, cos

anonymous · Answer

just thought it would be easier to solve, same answer in either case?

anonymous · Answer

is the answer even plausible?

loser66 · Answer

oh yea, you have to, sorry friend,

loser66 · Answer

so it turns $\dfrac{cosx}{1-sinx}$from 0 to $\tfrac{\pi}{2}$right?

anonymous · Answer

actually, cos(x)/(cos(x)-sin(x))

loser66 · Answer

$$\frac {cos \tfrac{\pi}{2}}{1-sin\frac{\pi}{2}}- \frac{cos0}{1-sin0}$$

loser66 · Answer

=-1

anonymous · Answer

but is that a real answer?

anonymous · Answer

there's an asymptot in the middle if we graph the function

loser66 · Answer

hehe... you got it before me, what do you mean by "real answer"? is there any fake answer?

loser66 · Answer

what is the original function? you just ask for taking integral, how can I know what happen to the original one? Even I know the original function, I may NOT know what happen with it.