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##\displaystyle \int \dfrac{1}{1+sin(\theta)}\ d\theta ##
I got ## -2(1+tan(\frac{\theta}{2}))^{-1} ## however wolframalpha got a different result than me, could anyone double check this for me as I've been over it a lot of times and can't get the wolframalpha equivalent
I got ## -2(1+tan(\frac{\theta}{2}))^{-1} ## however wolframalpha got a different result than me, could anyone double check this for me as I've been over it a lot of times and can't get the wolframalpha equivalent