Here,
#I=int1/(sinx+cosx)dx#
Let,
#tan(x/2)=t=>sec^2(x/2)*1/2dx=dt#
#=>dx=(2dt)/sec^2(x/2)=(2dt)/(1+tan^2(x/2))=(2dt)/(1+t^2#
also, #sinx=(2tan(x/2))/(1+tan^2(x/2))=(2t)/(1+t^2)#
and #cosx=(1-tan^2(x/2))/(1+tan^2(x/2))=(1-t^2)/(1+t^2)#
So,
#I=int1/((2t)/(1+t^2)+(1-t^2)/(1+t^2))xx(2dt)/(1+t^2#
#=int2/(2t+1-t^2)dt#
#=2int1/(2-t^2+2t-1)dt#
#=2int1/((sqrt2)^2-(t-1)^2)dt#
#=2xx1/(2sqrt2)ln|(t-1+sqrt2)/(t-1-sqrt2)|+c#
Subst. back , #t =tan(x/2)#
#I=1/sqrt2ln|(tan(x/2)+sqrt2-1)/(tan(x/2)-sqrt2-1)|+c#
Note: For typing simplicity #x# is taken in place of #theta#.