What is \int\frac{(\cos x)^{1.5}-(\sin x)^{1.5}}{\sqrt{\sin x\cdot \cos x}}\,dx equal to ?
- A. \sqrt{\sin x}-\sqrt{\cos x}+c
- B. \sqrt{\sin x}+\sqrt{\cos x}+c
- C. 2\sqrt{\sin x}+2\sqrt{\cos x}+c ✓
- D. \frac{1}{2}\sqrt{\sin x}+\frac{1}{2}\sqrt{\cos x}+c
Correct Answer: C. 2\sqrt{\sin x}+2\sqrt{\cos x}+c
Explanation
Divide each term in the numerator by the denominator: \frac{\cos^{1.5} x}{\sin^{0.5} x \cos^{0.5} x} - \frac{\sin^{1.5} x}{\sin^{0.5} x \cos^{0.5} x} = \frac{\cos x}{\sqrt{\sin x}} - \frac{\sin x}{\sqrt{\cos x}}. Integrating these terms: \int \frac{\cos x}{\sqrt{\sin x}} \,dx = 2\sqrt{\sin x} and \int \frac{-\sin x}{\sqrt{\cos x}} \,dx = 2\sqrt{\cos x}. Adding them gives 2\sqrt{\sin x} + 2\sqrt{\cos x} + c.
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