∫ e^x cot (e^x) dx = ? - tan e^x + c ... or is it another solution?

These were other possible book solutions............








a. e^x sin (e^x) + c





b. -e^x cos (e^x) + c





c. - cos e^x + c





d. sin e^x + c





e. - tan e^x + c





f. ln (sin e^x) + c





g. ln (sec e^x) + c





h. none of these

∫ e^x cot (e^x) dx = ? - tan e^x + c ... or is it another solution?
∫ e^x cot (e^x) dx





use u=e^x then du/dx = e^x


so the integral is now,


∫ cot u du = ln Isin uI +c = ln Isin e^xI + c


where c is an arbitrary constant of integration.


Hope this helps!
Reply:f.ln(sin e^x)
Reply:None are right since the tangent of a variable is not cosignable
Reply:e^x=u


e^x dx=du





integral( cot(u)du) =?


=ln|sinu|


=ln|sin(e^x)|+C
Reply:Change e^x = z so e^xdx =dz and your integral becomes





Int (coz/senz)dz = ln I sen z I+c = ln Isin e^x I





In the answer f) the absolute value is missing so this answer is partially right
Reply:answer would be ln (sin e^x)+c not would be i am sure
Reply:since it is an indefinite integration there exist more than one solution for a given question