Question

The value of integral $\int{e^x}(\frac{cosx+sinx}{cos^2x})dx$

• A. $e^{x} (sin \, x-sin\, x\, cos\, x)+C$
• B. $e^{x} (sin\,x \, cos\, x)+C$
• C. $e^xsecx+c$
• D. $e^{x} (sin \, x+cos \, x)+ C$

Answer 1

Answer: (C)

$e^xsecx+c$

Answer 2

The correct answer is C:$e^xsecx+c$
I=$\int{e^x}(\frac{cosx+sinx}{cos^2x})dx$
$I=\int{e^x}(\frac{1}{cosx}+tanx.secx)dx$
$I=\int{e^x}(secx+secx.tanx)dx$
Here, $f(x)=secx \therefore f'(x)=secx.tanx$
$\therefore I=\int{e^x}[f(x)+f'(x)]dx$
$I=e^xsecx+c$