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) \[\frac{1}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}\]
Correct Answer: A
Solution :
Given, \[\int{f(x)}\sin x\cos xdx=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\] \[\log [f(x)]+c\] On differentiating both sides, we get \[f(x)\sin x\cos x=\frac{d}{dx}\left( \frac{\log [f(x)]}{2({{b}^{2}}-{{a}^{2}})}+c \right)\] \[\Rightarrow \] \[f(x)\sin x\cos x=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\cdot \frac{1}{f(x)}f(x)\] \[\Rightarrow \]\[2({{b}^{2}}-{{a}^{2}})\sin x\cos x=\frac{f(x)}{{{[f(x)]}^{2}}}\] On integrating both sides, we get \[\int{(2{{b}^{2}}\sin x\cos x}-2{{a}^{2}}\sin x\cos x)dx\] \[=\int{\frac{f(x)}{{{[f(x)]}^{2}}}dx}\] \[\Rightarrow \,\,-{{b}^{2}}{{\cos }^{2}}x-{{a}^{2}}{{\sin }^{2}}x=-\frac{1}{f(x)}\] \[\therefore \,\,\,f(x)=\frac{1}{{{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x}\]
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