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Answer
$$ \int \frac{{{\sin{{x}}}{\cos{{x}}}}}{{{\left({\cos{{x}}}\right)}^{{2}}}}/{\left({\sin{{x}}}+{1}\right)} \, d\,x = {{\sin x\,\ln \left(\sin x+1\right)+\ln \left(\sin x+1\right)- \sin x\,\ln \left(\sin x-1\right)-\ln \left(\sin x-1\right)+2 }\over{4\,\left(\sin x+1\right)}} $$
Explanation
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