Q. $\int_{}^{}\frac{dx}{\cos{x - \sin x}}$ is equal to
A) $\frac{1}{\sqrt{2}}\log{\left| \tan\left( \frac{x}{2} - \frac{\pi}{8} \right) \right| + c}$
B) $\frac{1}{\sqrt{2}}\log{\left| \cot\left( \frac{x}{2} \right) \right| + c}$
C) $\frac{1}{\sqrt{2}}\log{\left| \tan\left( \frac{x}{2} - \frac{3\pi}{8} \right) \right| + c}$
D) $\frac{1}{\sqrt{2}}\log\left| \tan{\left( \frac{x}{2} + \frac{3\pi}{8} \right) + c} \right|$
Explanation
(d)
$I = \int_{}^{}\frac{dx}{\cos{x - \sin x}} = \frac{1}{\sqrt{2}}\ \int_{}^{}\frac{dx}{\left( {\frac{1}{\sqrt{2}}\ cos}{x - {\frac{1}{\sqrt{2}}\ sin}x} \right)}$
$= \frac{1}{\sqrt{2}}\ \int_{}^{}\frac{dx}{\cos\left( x + \frac{\pi}{4} \right)}$
$= \frac{1}{\sqrt{2}}\int_{}^{}{\sec{\left( x + \frac{\pi}{4} \right)dx}}$
$= \frac{1}{\sqrt{2}}\log\left| \tan\left( \frac{\pi}{4} + \frac{x}{2} + \frac{\pi}{8} \right) \right| + c$
$= \frac{1}{\sqrt{2}}\log\left| \tan\left( \frac{x}{2} + \frac{3\pi}{8} \right) \right| + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.