Question

Evaluate the indefinite integrals...
(cot5x+csc5x)dx

Answer 1

cos(5x)/sin(5x) + 1/sin(5x)dx
= cos(5x) + 1/sin(5x)dx

u = cos(5x) + 1
du = -5sin(5x)dx
-du/5 = sin(5x)dx

-1/5 integral of u dx
= -(ln|cos(5x) + 1| + c)/5

waiting for someone to correct me

Answer 2

Looks Right to me actually

Answer 3

the sin(5x) is canceled out

Answer 4

in the denominator

also

https://www.wolframalpha.com/input/?i=%28cot5%2Bcsc5%29+%3D+%28cos%285%29+%2B+1%29%2Fsin%285%29

proof that your initial problem can be simplified to what I simplified it to

Answer 5

wait never mind I made a mistake

Answer 6

you almost correct.
actually i already have a correct answer for that problem.
My problem is I dont know the solution. :)

Answer 7

this is the correct answer = 1/5ln(1-cos5x)+C
will you show me the solutions of it? :)

Answer 8

cos(5x)/sin(5x) + 1/sin(5x)dx
=
cos(5x) + 1/sin(5x)dx

u = cos(5x) + 1
du = -5sin(5x)dx
-du/5sin(5x) = dx

So we have

-1/5 integral of udu/sin^(2)(5x)

write in terms of u

so


arccos(u-1)/ = 5x

so

-1/5 integral of udu/(1-cos^(2)(arcos(u-1))
=
-1/5 integral of udu/(1-(u-1)^(2))
=
-1/5 integral of udu/1-(u^(2)-2u+1)
=
-1/5 integral of udu/1-u^(2)+2u-1
=
-1/5 integral of udu/(-u^(2)+2u)
=
-1/5 integral of udu/1-(u^(2)-2u+1)
=
-1/5 integral of udu/(-u^(2)+2u)
=
-1/5 integral of du/(-u+2)

Now another round of substitution I get

s = -u+2
ds = -1du
-ds = du

thus

1/5 integral of ds/s
=
(1/5)ln|s| + c
sub back in
(1/5)ln|-u+2| + c
sub back in
(1/5)ln|-cos(5x) - 1 + 2| + c
=
(1/5)ln|-cos(5x) + 1| + c

Answer 9

there is your answer it is a very dirty integral if you ask me lol

Answer 10

nice one @Australopithecus. Thanks a lot for your solutions. :)

Answer 11

if you have any questions about my solution feel free to ask It is kind of messy