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

Question

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

Answer 1

∫ 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!

Answer 2

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

Answer 3

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Answer 4

None are right since the tangent of a variable is not cosignable

Answer 5

answer would be ln (sin e^x)+c not would be i am sure

Answer 6

since it is an indefinite integration there exist more than one solution for a given question

Answer 7

e^x=u
e^x dx=du

integral( cot(u)du) =?
=ln|sinu|
=ln|sin(e^x)|+C

Answer 8

f.ln(sin e^x)