Question

Let $f\left(x\right)=\frac{2si{n}^{2}x-1}{cosx}+\frac{cosx(2sinx+1)}{1+sinx}$ then $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx$ equals
(where c is the constant of integeration)

• A. $e^{x}tanx+c$
• B. $e^{x}cotx+c$
• C. $e^{x}cose{c}^{2}x+c$
• D. None of these

Answer 1

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

$\frac{cosx(1+2sinx)}{1+sinx}-\frac{co{s}^{2}x-si{n}^{2}x}{cosx}=\frac{co{s}^{2}x(1+2sinx)-(1+sinx)(co{s}^{2}x-si{n}^{2}x)}{cosx(1+sinx)}$
$=\frac{sinxco{s}^{2}x+si{n}^{2}x(1+sinx)}{cosx(1+sinx)}=\frac{sinx(1-sinx)+si{n}^{2}x}{cosx}=tanx$