Question

integral 1 divided by 1+sinx dx

Answer 1

Do you mean?
\[\int\limits_{}^{}1dx/(1+\sin(x))\]

Answer 2

yes please

Answer 3

\[\int\limits_{}^{}\frac{1}{1+\sin(x)} \cdot \frac{1-\sin(x)}{1-\sin(x)} dx=\int\limits_{}^{}\frac{1-\sin(x)}{\cos^2(x)} dx\]
Did you try this?

Answer 4

no

Answer 5

caan you please continue with it for me. I'm an amateur with these things

Answer 6

Break into two fractions

Answer 7

write in terms of sec(x) and tan(x) and you will see it is not too terrible

Answer 8

\[\int\limits(1-\sin x)dx/\cos ^2x\]
Dividing (cosx)^2 into numerator we get,
\[\int\limits (\sec ^2x-\tan x \sec x)dx \]
Each term can be integrated independently.
\[\tan x-\sec x+C\]