Question

THIS QUESTION NEEDS TO BE INTEGRATED......

plzz answer this i have board exams plzz​

Answer 1

$\underline{\textsf{Solution :}}$

$\mathrm{Now,\:\int \frac{e^{x}(1+sinx)}{1+cosx}dx}$

$\mathrm{=\int \frac{e^{x}dx}{1+cosx}+\int \frac{e^{x}sinx\:dx}{1+cosx}}$

$\mathrm{=\int \frac{e^{x}dx}{sin^{2}\frac{x}{2}+cos^{2}\frac{x}{2}+cos^{2}\frac{x}{2}-sin^{2}\frac{x}{2}}}$

$\mathrm{+\int \frac{e^{x}(2\:sin\frac{x}{2}\:cos\frac{x}{2})dx}{sin^{2}\frac{x}{2}+cos^{2}\frac{x}{2}+cos^{2}\frac{x}{2}-sin^{2}\frac{x}{2}}}$

$\mathrm{=\int \frac{e^{x}dx}{2\:cos^{2}\frac{x}{2}}+\int \frac{e^{x}(2\:sin\frac{x}{2}\:cos\frac{x}{2})}{2\:cos^{2}\frac{x}{2}}}$

$\mathrm{= \int e^{x} sec^{2}\frac{x}{2}\:dx + \int e^{x}tan\frac{x}{2}\:dx}$

$\small{\mathrm{=\frac{1}{2} e^{x}\int sec^{2}\frac{x}{2}\:dx-\frac{1}{2} \int \{\frac{d(e^{x})}{dx} \int sec^{2}\frac{x}{2}\:dx\} dx}}$

$\mathrm{+ \int e^{x} tan\frac{x}{2}\:dx+C}$

$\textsf{where C is integral constant}$

$\mathrm{=\frac{2}{2}e^{x}tan\frac{x}{2}-\frac{2}{2} \int e^{x}tan\frac{x}{2}\:dx}$

$\mathrm{+\int e^{x}tan\frac{x}{2}\:dx+C}$

$\mathrm{=e^{x}tan\frac{x}{2}-\int e^{x}tan\frac{x}{2}\:dx}$

$\mathrm{+\int e^{x}tan\frac{x}{2}\:dx+C}$

$\mathrm{=e^{x}tan\frac{x}{2}+C}$

$\to \boxed{\mathrm{\int \frac{e^{x}(1+sinx)}{1+cosx}dx =e^{x}tan\frac{x}{2}+C}}$