Question

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

• A. \(\frac{x \csc x}{x\cos x - \sin x} + \cot x + c\)
• B. \(\frac{x \csc x}{x\cos x - \sin x} - \cot x + c\)
• C. \(\frac{x \csc x}{x\cos x + \sin x} + \cot x + c\)
• D. \(\frac{x}{x\cos x - \sin x} - \cot x + c\)

Hint:

Recognize the derivative of a quotient and use integration by parts.

Answer 1

The Correct Option is B

Step 1: Rewrite the integrand.
Let I = \(\int\left(\frac{x}{x\cos x - \sin x}\right)^2 dx\).
We can rewrite the integral as: I = \(\int \frac{x^2}{(x\cos x - \sin x)^2} dx\)

Step 2: Use integration by parts.
Let u = x csc x and dv = \(\frac{x \sin x}{(x\cos x - \sin x)^2} dx\).
Then du = csc x - x csc x cot x dx and v = \(-\frac{1}{x\cos x - \sin x}\).
Using integration by parts, \(\int u dv = uv - \int v du\):
I = \(\int \frac{x^2}{(x\cos x - \sin x)^2} dx\)
Let's try to find the derivative of \(\frac{\sin x - x \cos x}{x \cos x - \sin x}\).
\(\frac{d}{dx} \left(\frac{\sin x - x \cos x}{x \cos x - \sin x}\right) = \frac{(x \cos x - \sin x)(\cos x - \cos x + x \sin x) - (\sin x - x \cos x)(-x \sin x - \cos x + \cos x)}{(x \cos x - \sin x)^2}\)
= \(\frac{(x \cos x - \sin x)(x \sin x) - (\sin x - x \cos x)(-x \sin x)}{(x \cos x - \sin x)^2}\)
= \(\frac{x^2 \sin x \cos x - x \sin^2 x + x \sin^2 x \cos x - x^2 \cos^2 x}{(x \cos x - \sin x)^2}\)
= \(\frac{x^2 \sin x \cos x - x^2 \cos^2 x}{(x \cos x - \sin x)^2}\)
= \(\frac{x^2 \cos x (\sin x - \cos x)}{(x \cos x - \sin x)^2}\)

Step 3: Consider the derivative of \(\frac{x \csc x}{x \cos x - \sin x}\).
Let f(x) = \(\frac{x \csc x}{x \cos x - \sin x}\).
Then f'(x) = \(\frac{(x \cos x - \sin x)(\csc x - x \csc x \cot x) - (x \csc x)(-x \sin x - \cos x + \cos x)}{(x \cos x - \sin x)^2}\)
= \(\frac{(x \cos x - \sin x)\left(\frac{1}{sin x} - \frac{x \cos x}{\sin^2 x}\right) + x^2 \csc x \sin x}{(x \cos x - \sin x)^2}\)
= \(\frac{\frac{x \cos x - \sin x}{\sin x} - \frac{x \cos x(x \cos x - \sin x)}{\sin^2 x} + x^2}{(x \cos x - \sin x)^2}\)
= \(\frac{x \cos x \sin x - \sin^2 x - x^2 \cos^2 x + x \sin x \cos x + x^2 \sin^2 x}{\sin^2 x (x \cos x - \sin x)^2}\)
= \(\frac{2x \cos x \sin x - \sin^2 x - x^2 \cos^2 x + x^2 \sin^2 x}{\sin^2 x (x \cos x - \sin x)^2}\)

Step 4: Consider the derivative of \(\frac{x \csc x}{x \cos x - \sin x} - \cot x\).
\(\frac{d}{dx} \left(\frac{x \csc x}{x \cos x - \sin x} - \cot x\right) = \frac{x^2}{(x \cos x - \sin x)^2}\)
Therefore, the integral is \(\frac{x \csc x}{x \cos x - \sin x} - \cot x + c\).