Question

Let $$ I_n = \int (\cos^n x + \sin^n x) \, dx,\quad \text{and} \quad \frac{n - 1}{n} I_{n - 2} = \frac{\sin x \cos x}{n} f(x) $$ Then, what is $ f(x) $?

• A. \( \cos^{n - 2}x + \sin^{n - 2}x \)
• B. \( \cos^{n - 2}x - \sin^{n - 2}x \)
• C. \( \frac{\cos^{n - 2}x - \sin^{n - 2}x}{n} \)
• D. \( \frac{\cos^{n - 2}x + \sin^{n - 2}x}{n} \)

Hint:

When working with powers of trigonometric functions, apply reduction formulas and compare recursive forms to extract pattern.

Answer 1

The Correct Option is B

Given: \[ I_n = \int (\cos^n x + \sin^n x) dx \] and: \[ \frac{n - 1}{n} I_{n - 2} = \frac{\sin x \cos x}{n} f(x) \] We know that: \[ \int \cos^n x \, dx \Rightarrow \text{use reduction formula: } \int \cos^n x \, dx = \frac{\cos^{n-1}x \sin x}{n} + \frac{n - 1}{n} \int \cos^{n - 2}x \, dx \] Similarly for \( \sin^n x \). So: \[ I_n = \int \cos^n x \, dx + \int \sin^n x \, dx \Rightarrow \frac{n - 1}{n} I_{n - 2} = \frac{\sin x \cos x}{n} f(x) \] So the integrand must be: \[ f(x) = \cos^{n - 2}x - \sin^{n - 2}x \] Because multiplying by \( \sin x \cos x \) gives parts of the integration by parts or recursive form.