Question

\(\int \left( \frac{1}{x^2} + \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \right) dx =\)

• A. \(\frac{(\sin x - \cos x)x - \sin x \cos x}{x \sin x \cos x} + c\)
• B. \(\sec x - \csc x + c\)
• C. \(\frac{1}{x} \frac{\sin x - \cos x}{\sin^2 x \cos^2 x} + c\)
• D. \(\frac{(\sin x - \cos x)x - \sin x + \cos x}{x(\sin x + \cos x)} + c\)

Hint:

Always break down fractions with sums in the numerator (\(\frac{A+B}{C} = \frac{A}{C} + \frac{B}{C}\)) to simplify integrals involving trigonometric powers.

Answer 1

The Correct Option is A

Step 1: Simplify the Integrand:
Let the integral be \(I\). \[ I = \int \frac{1}{x^2} dx + \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} dx \] First part: \[ \int x^{-2} dx = -\frac{1}{x} \] Second part: Split the numerator: \[ \frac{\sin^3 x}{\sin^2 x \cos^2 x} + \frac{\cos^3 x}{\sin^2 x \cos^2 x} = \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} \] \[ = \sec x \tan x + \csc x \cot x \] Integrating the trigonometric part: \[ \int (\sec x \tan x + \csc x \cot x) dx = \sec x - \csc x \] Total Integral: \[ I = -\frac{1}{x} + \sec x - \csc x + c \] \[ I = \frac{1}{\cos x} - \frac{1}{\sin x} - \frac{1}{x} + c \]
Step 2: Match with Options:
Let's simplify Option (A): \[ \frac{(\sin x - \cos x)x - \sin x \cos x}{x \sin x \cos x} \] Split the fraction: \[ \frac{x(\sin x - \cos x)}{x \sin x \cos x} - \frac{\sin x \cos x}{x \sin x \cos x} \] \[ = \frac{\sin x - \cos x}{\sin x \cos x} - \frac{1}{x} \] \[ = \frac{\sin x}{\sin x \cos x} - \frac{\cos x}{\sin x \cos x} - \frac{1}{x} \] \[ = \frac{1}{\cos x} - \frac{1}{\sin x} - \frac{1}{x} \] \[ = \sec x - \csc x - \frac{1}{x} \] This matches our calculated result exactly.