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) \[{{a}^{2}}{{\cos }^{2}}x-{{b}^{2}}si{{n}^{2}}x+C\]
Correct Answer: A
Solution :
[a] \[\int{f(x)\sin \,\,x\,\,\cos \,\,x\,\,dx=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log (f(x))+C}\] Therefore \[f(x)\sin x\cos x=\frac{1}{2({{b}^{2}}-{{a}^{2}})}.\frac{1}{f(x)}f'(x)\] [by differentiating both the sides] \[\Rightarrow 2({{b}^{2}}-{{a}^{2}})\sin \,\,x\,\,\cos \,\,x=\frac{f'(x)}{{{\left( f(x) \right)}^{2}}}\] \[\int{(2{{b}^{2}}\sin x\cos x-2{{a}^{2}}\sin x\cos x)dx=\int{\frac{f'(x)}{{{\left( f(x) \right)}^{2}}}dx}}\][by integrating both the sides] \[\Rightarrow -{{b}^{2}}{{\cos }^{2}}x-{{a}^{2}}{{\sin }^{2}}x-C=-\frac{1}{f(x)}\]
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