Question

Evaluate ${\int }_0^{\pi /2}{e}^x\left(\frac{1+sinx}{1+cosx}\right)d{x}^{}$

• A. $\pi /4$
• B. $0$
• C. $e^{\pi /2}$
• D. $e^{\pi /2}-1$

Answer 1

The correct option is C $e^{\pi /2}$

$I={\int }_0^{\frac{\pi }{2}}{e}^x\left(\frac{1+sinx}{1+cosx}\right)dx$
$={\int }_0^{\frac{\pi }{2}}{e}^x(\frac{1}{1+cosx}+\frac{sinx}{1+cosx})dx$
Let $f\left(x\right)=\frac{sinx}{1+cosx}\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{1}{1+cosx}$
Using $\int e^x(f\left(x\right)+f^{{}^{\prime }}\left(x\right))=e^{x}f\left(x\right)+c$
We get
$I={\left[e^{x}f\left(x\right)\right]}_0^{\frac{\pi }{2}}={\left[e^x\frac{sinx}{1+cosx}\right]}_0^{\frac{\pi }{2}}=e^{\frac{\pi }{2}}$