Question : Integrate [x/(1+(sin a*sin x))] from 0 to pi
My first thought was to apply integrate f(x) dx= f(a-x) dx method
Which simplified the integral into;
2I = integrate [pi/(1+(sin a*sin x))] dx , cancelling out x
Then I made the integral into the form of the following by making necessary changes;
2I* sin a = integrate [pi*((sin a*cos x)*sec x)/(1+ (sin a* sin x))]
Then integration by parts,
2I*sin a = pi*(ln| 1+(sina* cos x)|*sec x ) - integrate [pi*(ln|(1 +(sin a*sin x))|*(sec x*tan x) dx
How do I simplify ln|(1+(sin a*sin x))| to finish integrating this?
I don't think I can help you on this but hopefully Ms. Sue will come soon.
To simplify the expression ln|(1 + (sin a * sin x))|, you can use the logarithmic identity:
ln(1 + u) = ln|u + sqrt(1 + u^2)| - ln|sqrt(1 + u^2) - u|,
where u = sin a * sin x.
Applying this identity to your expression, we have:
ln|(1 + (sin a * sin x))| = ln|(sin a * sin x) + sqrt(1 + (sin a * sin x)^2)| - ln|sqrt(1 + (sin a * sin x)^2) - (sin a * sin x)|.
Now you can substitute this expression back into your integral:
2I * sin a = pi * (ln|(sin a * cos x) * sec x|) - pi * (ln|(sin a * sin x) + sqrt(1 + (sin a * sin x)^2)| - ln|sqrt(1 + (sin a * sin x)^2) - (sin a * sin x)| * (sec x * tan x) dx.
From here, you can integrate each term using the appropriate integration techniques.