Question

need help with this definite integral : (posting the question below)

Answer 1

solve : \[\frac{ dy }{ dx } = \sin (x+y) + \cos (x+y) -1 \]
the answer given is : \[\log_{} [cosec (x+y+ \frac{\pi}{4}) - \cot (x+y+ \frac{\pi}{4}) ] = \sqrt{2}x +c\]

Answer 2

i took x+y =z
the equation then became \[\frac{ dz }{ dx } = \sin z - \cos z\]
or \[\int\limits \frac {dz}{\sin z + \cos z} = \int\limits dx \]
i then substituted sin z and cos z in terms of tan x/2 , but not getting the answer

Answer 3

if z=x+y
then z'=1+y'
so you would have z'-1=sin(z)+cos(z)-1
so z'=sin(z)+cos(z) but I think you made that correct in the line after
tan(x/2)=u
sin(x)=2sin(x/2)cos(x/2)=2u/(1+u^2)
cos(x)=cos^2(x/2)-sin^2(x/2)=1/(1+u^2)-u^2/(1+u^2)=(1-u^2)/(1+u^2)
so we have
\[\int\limits_{}^{}\frac{1}{\frac{2u}{1+u^2}+\frac{1-u^2}{1+u^2}} du =\int\limits_{}^{}\frac{1+u^2}{2u+(1-u^2)} du\]
Is that what you got?

Answer 4

for the left hand side anyways

Answer 5

i got \[\int\limits\limits_{}^{} \frac{\sec^2 \frac{z}{2}}{1-\tan^2 \frac{z}{2}-2 \tan \frac{z}{2}} dz\]