The value of $\int_0^1 \frac{(\sqrt{1-x^2})}{x^2} dx$ is equal to
Step-by-Step Solution
Key Concept: The substitution $x = \cos 6\theta$ reduces the integral to a trigonometric form that can be evaluated using standard identities
Let $t = \sqrt{x}$, so $x = \cos 6\theta$ and $dx = 6\cos 6\theta d\theta$. Then $I = \int_{1/2}^{\pi/6} \frac{(\sin 6\theta)}{\cos^2 6\theta} 6\cos 6\theta d\theta = \int_{1/2}^{\pi/6} \tan^2 \theta \, d\theta = \int_{1/2}^{\pi/6} \cos^2(\cos^2 \theta - 1) d\theta = \int_{1/2}^{\pi/6} \cos^2 \theta \cos^2 \theta d\theta - \int \cos^2 \theta d\theta = -\left[\frac{u d\theta}{4}\right]_{1/2}^{\pi/6} - \int(\cos^2 \theta - 1) d\theta = \sqrt{3} + [0 - \sqrt{3}] + (\frac{7}{2} - \frac{2}{2}) = \frac{7}{2}$.
Correct Answer: 3