Please help with this integral !!! sinhxcos^2x - (cosx)/(cosx+sinx) in the borders -pi/2 to pi/2

Question

turingtest · Answer

well this gets pretty crazy, but I think it's on the right track...

anonymous · Answer

man you are saving me ... !!!!! =] =]

turingtest · Answer

$$\sinh x\cos^2x-\frac{\cos x}{\cos x+\sin x}=\sinh\cos^2x-\frac{ 1}{1+\tan x}$$let$$u=\tan x$$$$du=\sec^2xdx$$and note that $$\cos^2 x=\frac1{1+u^2}$$$$dx=\frac{du}{1+u^2}$$

anonymous · Answer

how did you transform to 1/1+tanx ?

turingtest · Answer

I multiplied the fraction by secx/secx

anonymous · Answer

ohh got it... nice !!! =]

turingtest · Answer

also note that$$x=\tan^{-1}u$$so the expression becomes$$\frac{\sinh(\tan^{-1}u)}{u^2+1}+\frac1{(1+u)(1+u^2)}du$$notice that the first part can be handled by another simple sub, because the derivative of arctanu is 1/(u^2+1)

the second part would be partial fractions...

turingtest · Answer

should be a minus sign between them*

anonymous · Answer

man you nailed it .. i think that is the right way !!!

turingtest · Answer

Thanks, I hope so
good luck :D

anonymous · Answer

thank you !! pleasure having you here =]