Question

The integral $I=\int e^x\left(\frac{1+sinx}{1+cosx}\right)dx=e^{x}f\left(x\right)+C$ (where, $C$ is the constant of integration). Then, the range of $y=f\left(x\right)$ (for all $x$ in the domain of $f\left(x\right)$ ) is

• A. $\left[-1,1\right]$
• B. $\left(-\in fty,\in fty\right)$
• C. $\left(-1,1\right)$
• D. $[0,\infty )$

Answer 1

Answer: (B)

The given integral $I=\int e^x\left(\frac{1+2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\cos }^2\frac{x}{2}}\right)dx$
$=\int e^x\left(\frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)+\tan \left(\frac{x}{2}\right)\right)dx$
$=e^x\cdot \tan \left(\frac{x}{2}\right)+C\left\{\because \int e^x\left(f(x)+f^{\prime }(x)\right)\left(dx=e^{x}f(x)+C\right)\right\}$
Thus, $f(x)=\tan \left(\frac{x}{2}\right)$ whose range is $(-\infty ,\infty )$