Question

Integrate the function: \(e^x(\frac{1+sinx}{1+cosx})\)

Answer 1

The correct answer is: \(e^xtan\frac{x}{2}+C\)
\(e^x(\frac{1+sinx}{1+cosx})\)
\(=\bigg(\frac{e^x\frac{sin^2x}{2}+\frac{cos^2x}{2}+2sin\frac{x}{2}cos\frac{x}{2}}{2cos^2\frac{x}{2}}\bigg)\)
\(=\frac{e^x(sin\frac{x}{2}+cos\frac{x}{2})^2}{2cos^2\frac{x}{2}}\)
\(=\frac{1}{2}e^x.\bigg(\frac{(sin\frac{x}{2}+cos\frac{x}{2})}{(cos\frac{x}{2})}\bigg)^2\)
\(=\frac{1}{2}e^x[tan\frac{x}{2}+1]^2\)
\(=\frac{1}{2}e^2(1+tan\frac{x}{2})^2\)
\(=\frac{1}{2}e^x[1+tan^2\frac{x}{2}+2tan\frac{x}{2}]\)
\(=\frac{1}{2}e^x[sec^2\frac{x}{2}+2tan\frac{x}{2}]\)
\(\frac{e^x(1+sinx)dx}{(1+cosx)}=e^x[\frac{1}{2}sec^2\frac{x}{2}+tan\frac{x}{2}]...(1)\)
Let \(tan\frac{x}{2}=ƒ(x) \,\,ƒ'(x)=\frac{1}{2}sec^2 \frac{x}{2}\)
It is known that,\(∫e^x[ƒ(x)+ƒ'(x)]dx=e^x ƒ(x)+C\)
From equation(1),we obtain
\(∫\frac{e^x(1+sinx)}{(1+cosx)}dx=e^xtan\frac{x}{2}+C\)