Question

$\int 2e^x\frac{\sqrt{1+sin2x}}{1+cos2x}dx=$

• A. $e^{x}secx+c$
• B. $e^x$ tan $x+c$
• C. $e^x$ secx tan $x+c$
• D. $e^x$ cosec $x+c$

Answer 1

The correct option is A $e^{x}secx+c$

$\int 2e^x\frac{\sqrt{1+sin2x}}{1+cos2x}dx=\int 2e^x\frac{sinx+cosx}{2co{s}^{2}x}dx$

$\int e^x(tanxsecx+secx)dx$

it is in the form of

$\int e^x\left(f\right(x)+f^1(x\left)\right)dx$

$=e^{x}f\left(x\right)+c$

so

$\int e^x(tanxsecx+secx)dx=e^{x}secx+c$

$so,\int 2e^x\frac{\sqrt{1+sin2x}}{1+cos2x}dx=e^{x}secx+c$

$whilexϵ(-\pi /4,3\pi /4)$