Q. $\int_{0}^{\frac{\pi}{6}}\frac{\sin x}{\cos^{3}x}\ dx$ is equal to
A) $\frac{2}{3}$
B) $\frac{1}{6}$
C) 2
D) $\frac{1}{3}$
Explanation
(b)
Let $I = \int_{0}^{\pi/6}{\frac{\sin x}{\cos^{3}x}dx}$
Put $\cos{x = t\ \Rightarrow \ - \sin{x\ dx = dt}}$
$\therefore I = - \int_{1}^{\sqrt{3}/2}\frac{1}{t_{3}} = \left\lbrack \frac{1}{2t^{2}} \right\rbrack_{1}^{\sqrt{3}/2} = \frac{1}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.