Question

Calculate for following integral : \[\int\limits cosecx dx\] by substituting t=tan(x/2)

Answer 1

\[ \int \frac{1}{sin(x)}dx\]
\[ \int \frac{1}{ \frac{2t}{1+t^2}}dt = \int \frac{1+t^2}{2t}dt = \int \frac{1}{2t} + \frac{t^2}{2t} dt\]

Answer 2

http://en.wikipedia.org/wiki/Tangent_half-angle_substitution

Answer 3

\[ \frac{1}{2} \int \frac{1}{t} dt + \int \frac{t}{2}dt\]

Answer 4

\[\frac{1}{2} \int \frac{1}{t} dt + \frac{1}{2}\int tdt\]

Answer 5

Ok, so then you get 1/2lnt + 1/4t^2?

Answer 6

hmm..something is wrong.

Answer 7

with my answer?

Answer 8

with mine..

Answer 9

I think I know, we should have substituted 2dt/1+t\[^2\] in for dx, as it says on the wiki page

Answer 10

im sooo stupiddd lol!
i forgot about \[ dx = \frac{2}{1+t}\] lol

Answer 11

haha don't worry, I've followed through with that and it works out

Answer 12

\[ \int \frac{1}{sinx} dx = \int \frac{ 1+t^2}{2t} * \frac{2}{1+t^2}\]
\[ \int \frac{1}{t} dt = ln|t| +c\]
better now hehe:D

Answer 13

yep exactly