If \[\int_{{}}^{{}}{f(x)\sin x\cos x\ dx=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log (f(x))}+c\], then \[f(x)=\]

\[\frac{1}{{{a}^{2}}{{\sin }^{2}}x-{{b}^{2}}{{\cos }^{2}}x}\]

\[\frac{1}{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}\]

\[\frac{1}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}\]

\[\frac{1}{{{a}^{2}}{{\cos }^{2}}x-{{b}^{2}}{{\sin }^{2}}x}\]

If \[\int_{{}}^{{}}{f(x)\sin x\cos x\ dx=\frac{1}{2({{b}^{2}}-{{a}^{2}

1.	If \[\int_{{}}^{{}}{f(x)\sin x\cos x\ dx=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log (f(x))}+c\], then \[f(x)=\]
A.	\[\frac{1}{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}\]
B.	\[\frac{1}{{{a}^{2}}{{\sin }^{2}}x-{{b}^{2}}{{\cos }^{2}}x}\]
C.	\[\frac{1}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}\]
D.	\[\frac{1}{{{a}^{2}}{{\cos }^{2}}x-{{b}^{2}}{{\sin }^{2}}x}\]
Answer» B. \[\frac{1}{{{a}^{2}}{{\sin }^{2}}x-{{b}^{2}}{{\cos }^{2}}x}\]

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If \[\int_{{}}^{{}}{f(x)\sin x\cos x\ dx=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log (f(x))}+c\], then \[f(x)=\]

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