If $$\int {f(x)} \sin x\cos xdx = {1 \over {2({b^2} - {a^2})}}\log (f(x)) + c$$, where c is the constant of integration, then f(x) is equal to
- A$${2 \over {({b^2} - {a^2})\sin 2x}}$$
- B$${2 \over {ab\sin 2x}}$$
- C$${2 \over {({b^2} - {a^2})\cos 2x}}$$
- D$${2 \over {ab\cos 2x}}$$
✓ Correct answer: C
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