Question

The value of definite integral $\int_{0}^{\pi/2} \frac{\cos x}{(\sin x + \sqrt{3} \cos x)^3} dx$ equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $\frac{1}{12}$

Answer 1

Answer: (C)

$\frac{1}{6}$

Answer 2

To evaluate the definite integral $I = \int_{0}^{\pi/2} \frac{\cos x}{(\sin x + \sqrt{3} \cos x)^3} dx$, we can use a substitution.

First, divide the numerator and the denominator by $\cos^3 x$: $\frac{\cos x}{(\sin x + \sqrt{3} \cos x)^3} = \frac{\frac{\cos x}{\cos^3 x}}{\frac{(\sin x + \sqrt{3} \cos x)^3}{\cos^3 x}} = \frac{\sec^2 x}{\left(\frac{\sin x}{\cos x} + \sqrt{3} \frac{\cos x}{\cos x}\right)^3} = \frac{\sec^2 x}{(\tan x + \sqrt{3})^3}$ The integral becomes: $I = \int_{0}^{\pi/2} \frac{\sec^2 x}{(\tan x + \sqrt{3})^3} dx$ Now, let $u = \tan x + \sqrt{3}$. Then, $du = \sec^2 x \, dx$.

We need to change the limits of integration: When $x = 0$, $u = \tan(0) + \sqrt{3} = 0 + \sqrt{3} = \sqrt{3}$. When $x = \frac{\pi}{2}$, as $x \to \frac{\pi}{2}^-$, $\tan x \to \infty$, so $u \to \infty$.

The integral transforms to: $I = \int_{\sqrt{3}}^{\infty} \frac{1}{u^3} du$ This is an improper integral. Evaluating it: $I = \int_{\sqrt{3}}^{\infty} u^{-3} du = \left[ \frac{u^{-3+1}}{-3+1} \right]_{\sqrt{3}}^{\infty} = \left[ -\frac{1}{2u^2} \right]_{\sqrt{3}}^{\infty}$ Applying the limits: $I = \lim_{b \to \infty} \left( -\frac{1}{2b^2} \right) - \left( -\frac{1}{2(\sqrt{3})^2} \right)$ $I = 0 - \left( -\frac{1}{2 \cdot 3} \right) = 0 - \left( -\frac{1}{6} \right) = \frac{1}{6}$