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Question: If \(\int_{}^{}{f(x)}\)sin x cos x dx = \(\frac{1}{2(b^{2} - a^{2})}\)ln f(x) + c, then f(x) is –...

If ∫f(x)\int_{}^{}{f(x)}sin x cos x dx = 12(b2−a2)\frac{1}{2(b^{2} - a^{2})}ln f(x) + c, then f(x) is –

A

1asin⁡x+bcos⁡x\frac{1}{a\sin x + b\cos x}

B

1a2sin⁡2x+b2cos⁡2x\frac{1}{a^{2}\sin^{2}x + b^{2}\cos^{2}x}

C

1a2sin⁡x+b2cos⁡x\frac{1}{a^{2}\sin x + b^{2}\cos x}

D

1asin⁡2x+bcos⁡2x\frac{1}{a\sin^{2}x + b\cos^{2}x}

Answer

1a2sin⁡2x+b2cos⁡2x\frac{1}{a^{2}\sin^{2}x + b^{2}\cos^{2}x}

Explanation

Solution

Given ∫f(x)\int_{}^{}{f(x)}sin x cos x dx = 12(b2−a2)\frac{1}{2(b^{2} - a^{2})} ln f(x) + c

Differentiating both sides w.r.t.x then

f(x) sin x cos x = 12(b2−a2)\frac{1}{2(b^{2} - a^{2})}. f′(x)f(x)\frac{f'(x)}{f(x)}

Ž 2(b2 – a2) sin x cos = f′(x){f(x)}2\frac{f'(x)}{\{ f(x)\}^{2}}

Ž 2b2 sin x cos x – 2a2 sin x cos x = f′(x){f(x)}2\frac{f'(x)}{\{ f(x)\}^{2}}

Integrating both side w.r.t.x we get

–b2 cos2 x – a2 sin2 x = – 1f(x)\frac{1}{f(x)}

Or f(x) = 1(a2sin⁡2x+b2cos⁡2x)\frac{1}{(a^{2}\sin^{2}x + b^{2}\cos^{2}x)}