1. ## calculate

2. $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+sinx}dx=$

$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+sinx}\left(\ frac{1-sinx}{1-sinx}\right)dx$

$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sinx}{cos^2x}dx-\int_{0}^{\frac{\pi}{2}}\frac{sin^2x}{cos^2x}dx$

the first one substitute u=cosx the other tan^2x+1=sec^x .... tan^2x=sec^2-1 and now the integrate for sec^2x =tanx and integrate of -1 =-x

thats all

3. they cannot multiply and divide on 1-sinx; because 1-sinx is can equal to zero
thankyou

4. ## help

help: let us pose t= Tan(x/2)

5. you know the answer right ??

but if you derive my answer you will get the integral I try it I think there is no problem if I multiply with 1-sinx

or if the integral have limits then we should look at the interval [0,pi/2]
I cant multiply with 1-sinx or 1-cosx or 1-tanx .....

6. Originally Posted by Amer
you know the answer right ??

but if you derive my answer you will get the integral I try it I think there is no problem if I multiply with 1-sinx

or if the integral have limits then we should look at the interval [0,pi/2]
I cant multiply with 1-sinx or 1-cosx or 1-tanx .....
remark : 1-sinx=0
sinx=1
x=pi/2?????

7. I know I ask if the integral have not limits then I can multiply with 1-sinx