Question

Evaluate :$\int \left(\frac{e^x[\sin x+\cos x]}{1-{\sin }^{2}x}\right)dx$

• A. $e^x\cos ecx+c$
• B. $e^x+cotx+c$
• C. $e^{x}secx+c$
• D. $e^x\tan x+c$

Answer 1

The correct option is C

$e^{x}secx+c$

Explanation for the correct option

Finding $\int \left(\frac{e^x[\sin x+\cos x]}{1-{\sin }^{2}x}\right)dx$

Solving the integral,

$\begin{array}{rcl}& =& \int \frac{e^x[\sin x+\cos x]}{(1–si{n}^{2}x)}dx\\ & =& \int \frac{e^x\left[\sin x+\cos x\right]}{co{s}^{2}x}dx[\hspace{0.17em}1={\cos }^2\left(x\right)+{\sin }^2\left(x\right)]\\ & =& \int e^x\left(\frac{\sin x}{{\cos }^{2}x}+\frac{\cos x}{{\cos }^{2}x}\right)dx\\ & =& \int e^x(secx\times tanx+secx)dx\\ & =& \left[e^{x}secx\right]+c\left[\because \int e^x\left(f\left(x\right)+f\text{'}\left(x\right)\right)=e^{x}f\left(x\right)+c\right]\end{array}$

Hence, the correct answer is Option (C)