Question

The value of $\displaystyle\int^{\frac{\pi}{2}}_{\frac{\pi}{6}}\frac{\cot x}{\sin x}dx$ is equal to

• A. $\frac{-1}{2}$
• B. \(\frac{\1}{2}\)
• C. $\frac{-3}{2}$
• D. $\frac{3}{2}$
• E. 1

Answer 1

Answer: (E)

We are asked to evaluate the integral $$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cot x}{\sin x} \, dx.$$ First, express the integrand in a simpler form: $$\frac{\cot x}{\sin x} = \frac{\cos x}{\sin^2 x}.$$ Now, the integral becomes: $$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cos x}{\sin^2 x} \, dx.$$ Next, let’s make the substitution $u = \sin x$, so that $du = \cos x \, dx$. The limits of integration change accordingly: when $x = \frac{\pi}{6}$, $u = \sin \frac{\pi}{6} = \frac{1}{2}$; and when $x = \frac{\pi}{2}$, $u = \sin \frac{\pi}{2} = 1$. The integral becomes: $$I = \int_{\frac{1}{2}}^{1} \frac{du}{u^2}.$$ Now, integrate $\frac{1}{u^2}$: $$I = \left[ -\frac{1}{u} \right]_{\frac{1}{2}}^{1} = -\frac{1}{1} + \frac{1}{\frac{1}{2}} = -1 + 2 = 1.$$

The correct option is (E) : $1$