Question

evaluate each integral (csc^2 * t) / (cot t) dx

Answer 1

write it by usgin equation plz so we can read : )

Answer 2

it can be written 1/sin(x)cos(x)
2csc(2x)
now take out integral
ln(csc2x-cot2x)+c

Answer 3

\[\frac{1}{sin^2t}*\frac{sin t}{cost}\implies \frac{1}{sint *cost}\]\[t = sinx\]\[dx=\frac{dt}{cosx}\]
\[\int\frac{1}{t(1-t^2)}dt\]

Answer 4

Now Integrate it using partial fraction

Answer 5

I used this idetity
sin(2x)=2sinxcosx

Answer 6

okay so for some reason this site is very slow and myequation was deleted :(

Answer 7

since d/dt(cot(t))=-csc^2t
\[\int\limits_{}^{}-1/\cot(t)d(\cot(t))=\int\limits_{}^{}\csc^2(t)/\cot(d)d\]
so since the derivative of cot(t) is sitting in the integral
you can let "u" be cot(t)
INT 1/u=ln(u)=-ln(cot(t))
you add a (-) sign because the d/dt(cot(t))=-csc^2t
answer=-ln(cot(t)) +C

Answer 8

you meant dt at the top right not dx?

Answer 9

dt yes I messed up.

Answer 10

There’s more than one way to skin a cat : )

Answer 11

yes, do you understand what i mean when i say
\[\int\limits_{}^{}\csc^2(t)/\cot(t)dt=\int\limits_{}^{}-1/\cot(t)d(\cot(t))dt\]