If \[\int{f(x)\sin x\cos x\,\,dx=\frac{1}{2({{b}^{2}}-{{a}^{2}})}{{\log }_{e}}(f(x))+A,}\]\[b\ne \pm a,\] then \[{{\{f(x)\}}^{-1}}\] is equal to

\[{{a}^{2}}{{\sin }^{2}}x-{{b}^{2}}{{\cos }^{2}}x+C\]

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

\[{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}si{{n}^{2}}x+C\]

\[{{a}^{2}}{{\cos }^{2}}x-{{b}^{2}}si{{n}^{2}}x+C\]

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

1.	If \[\int{f(x)\sin x\cos x\,\,dx=\frac{1}{2({{b}^{2}}-{{a}^{2}})}{{\log }_{e}}(f(x))+A,}\]\[b\ne \pm a,\] then \[{{\{f(x)\}}^{-1}}\] is equal to
A.	\[{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x+C\]
B.	\[{{a}^{2}}{{\sin }^{2}}x-{{b}^{2}}{{\cos }^{2}}x+C\]
C.	\[{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}si{{n}^{2}}x+C\]
D.	\[{{a}^{2}}{{\cos }^{2}}x-{{b}^{2}}si{{n}^{2}}x+C\]
Answer» B. \[{{a}^{2}}{{\sin }^{2}}x-{{b}^{2}}{{\cos }^{2}}x+C\]

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If \[\int{f(x)\sin x\cos x\,\,dx=\frac{1}{2({{b}^{2}}-{{a}^{2}})}{{\log }_{e}}(f(x))+A,}\]\[b\ne \pm a,\] then \[{{\{f(x)\}}^{-1}}\] is equal to

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