Q. The anti-derivative of $f(x) = \frac{1}{3 + 5\sin{x + 3\cos x}}$, whose graph passes through the point (0, 0) is
A) $\frac{1}{5}\left( \log\left| 1 - \frac{5}{3}\tan{(x/2)} \right| \right)$
B) $\frac{1}{5}\left( \log\left| 1 + \frac{5}{3}\tan{(x/2)} \right| \right)$
C) $\frac{1}{5}\left( \log\left| 1 + \frac{5}{3}\cot{(x/2)} \right| \right)$
D) None of these
Explanation
(b)
$y = \int_{}^{}\frac{dx}{(3 + 5\sin{x + 3\cos{x)}}}$
$= \int_{}^{}{\frac{\sec^{2}{(x/2)}}{10\tan{(x/2) + 6}}dx}$
$= \frac{1}{5}\log{|5\tan{(x/2) + 3| + c}}$
$= \frac{1}{5}\log{\left| \frac{5}{3}\tan{(x/2) + 1|} \right| + c}$
This curve passes through (0, 0).
$\therefore c = 0$
Then, $y = \frac{1}{5}\log\left| 1 + \frac{5}{3}\tan\left( \frac{x}{2} \right) \right|$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.