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Mathematics Question on Definite Integral

If ∫ f(x) sin⁡ x cos⁡ x dx=12(b2−a2)log⁡f(x)+c\int \, f(x) \, \sin \, x \, \cos \, x \, dx = \frac{1}{2(b^2 - a^2)} \log f(x) + c, where c is the constant of integration , then f(x) =

A

2(b2−a2)sin⁡ 2x\frac{2}{(b^2 - a^2) \sin \, 2x}

B

2ab sin⁡ 2x\frac{2}{ab\, \sin \, 2x}

C

2(b2−a2)cos⁡ 2x\frac{2}{(b^2 - a^2) \cos \, 2x}

D

2ab cos⁡ 2x\frac{2}{ab \, \cos \, 2x}

Answer

2(b2−a2)cos⁡ 2x\frac{2}{(b^2 - a^2) \cos \, 2x}

Explanation

Solution

We have, ∫f(x)sin⁡xcos⁡xdx=12(b2−a2)log⁡(f(x))+c\int f(x) \sin x \cos x d x=\frac{1}{2\left(b^{2}-a^{2}\right)} \log (f(x))+c
⇒f(x)sin⁡xcos⁡x=12(b2−a2)⋅1f(x)⋅f′(x)\Rightarrow f(x) \sin x \cos x=\frac{1}{2\left(b^{2}-a^{2}\right)} \cdot \frac{1}{f(x)} \cdot f'(x)
⇒f(x)sin⁡2x=1b2−a2⋅f′(x)f(x)\Rightarrow f(x) \sin 2 x=\frac{1}{b^{2}-a^{2}} \cdot \frac{f'(x)}{f(x)}
⇒sin⁡2x=1b2−a2f′(x)(f(x))2\Rightarrow \sin 2 x=\frac{1}{b^{2}-a^{2}} \frac{f'(x)}{(f(x))^{2}}
⇒∫sin⁡2xdx=1b2−a2∫f′(x)(f(x))2dx\Rightarrow \int \sin 2 x d x=\frac{1}{b^{2}-a^{2}} \int \frac{f'(x)}{(f(x))^{2}} d x
⇒−cos⁡2x2=1b2−a2⋅(−1f(x))\Rightarrow \frac{-\cos 2 x}{2}=\frac{1}{b^{2}-a^{2}} \cdot\left(\frac{-1}{f(x)}\right)
⇒cos⁡2x(b2−a2)2=1f(x)\Rightarrow \frac{\cos 2 x\left(b^{2}-a^{2}\right)}{2}=\frac{1}{f(x)}
⇒f(x)=2(b2−a2)cos⁡2x\Rightarrow f(x)=\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}