- #1
Aresius
- 49
- 0
Hi I have a proof I'm doing
[tex]
\int \frac{1}{1+\sin(x)}dx
[/tex]
I know that the answer I'm looking for is
[tex]
\frac{\sin(x) - 1}{\cos(x)}
[/tex]
and then
[tex]
\tan(x) - \sec(x)
[/tex]
I have tried integration by parts making
[tex]
u = (1+\sin(x))^{-1}[/tex] and [tex]dv = dx
[/tex]
Eventually I get an answer that contains an ln and an unsolvable integral. I have been at this for 2 hours, can anyone give me a hint or a push in the right direction?
[tex]
\int \frac{1}{1+\sin(x)}dx
[/tex]
I know that the answer I'm looking for is
[tex]
\frac{\sin(x) - 1}{\cos(x)}
[/tex]
and then
[tex]
\tan(x) - \sec(x)
[/tex]
I have tried integration by parts making
[tex]
u = (1+\sin(x))^{-1}[/tex] and [tex]dv = dx
[/tex]
Eventually I get an answer that contains an ln and an unsolvable integral. I have been at this for 2 hours, can anyone give me a hint or a push in the right direction?
Last edited: