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) : \(1\)

Answer 2

We are asked to evaluate the definite integral: \[\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cot x}{\sin x} dx\]

We can rewrite the integrand as: \[\frac{\cot x}{\sin x} = \frac{\frac{\cos x}{\sin x}}{\sin x} = \frac{\cos x}{\sin^2 x}\]

So the integral becomes: \[\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cos x}{\sin^2 x} dx\]

Now, we use the substitution method. Let \(u = \sin x\). Then, the derivative of \(u\) with respect to \(x\) is: \[\frac{du}{dx} = \cos x\] \[du = \cos x \, dx\]

We also need to change the limits of integration. When \(x = \frac{\pi}{6}\), \(u = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\). When \(x = \frac{\pi}{2}\), \(u = \sin\left(\frac{\pi}{2}\right) = 1\).

Substitute \(u\) and \(du\) into the integral, along with the new limits of integration: \[\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cos x}{\sin^2 x} dx = \int_{\frac{1}{2}}^{1} \frac{1}{u^2} du = \int_{\frac{1}{2}}^{1} u^{-2} du\]

Integrate with respect to \(u\): \[\int_{\frac{1}{2}}^{1} u^{-2} du = \left[\frac{u^{-1}}{-1}\right]_{\frac{1}{2}}^{1} = \left[-\frac{1}{u}\right]_{\frac{1}{2}}^{1}\]

Evaluate the definite integral: \[\left[-\frac{1}{u}\right]_{\frac{1}{2}}^{1} = -\frac{1}{1} - \left(-\frac{1}{\frac{1}{2}}\right) = -1 - (-2) = -1 + 2 = 1\]

Therefore, the definite integral is: \[\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cot x}{\sin x} dx = 1\]