Question

Find the following integral.

Answer 1

which ?

Answer 2

lol hold on im writing it out now.

Answer 3

\[ ∫\frac{ cosx }{ 1-\cos ^{2}x } dx\]

Answer 4

can't you simplify the denominator ?
\(\sin^2 x+\cos^2x=1 \\ 1-\cos^2x=... ?\)

Answer 5

sin^2(x)

Answer 6

then i took out a sin from the bottom to make it cotx * 1/sinx

Answer 7

you have cos/sin^2
keep it like that only.
put u= sin x
du=.... ?

Answer 8

cos2x d(2x)

Answer 9

?? how ?
u= sin x dx
du will be just cos x dx
isn't it ?

Answer 10

im sorry i've got to run across campus now, ill be back in like 15 mins though!
thanks for your help so far.

Answer 11

and yes lol.

Answer 12

no problem :)

Answer 13

okay sorry about that im back now.

Answer 14

good to see you back :)
try u= sin x
du=... ?

Answer 15

cos x dx?

Answer 16

yes, so numerator = du
denominator = u^2
can you integrate that ?

Answer 17

i dont know how to integrate the bottom part.

Answer 18

you have \(\int du/u^2= \int u^{-2}du=..?\)
use the formula for integarl of x^n dx.....