A) \[\frac{2}{ab\,\,\cos \,\,2x}\]
B) \[\frac{2}{({{b}^{2}}-{{a}^{2}})\,\cos \,2x}\]
C) \[\frac{2}{ab\,\sin \,2x}\]
D) \[\frac{2}{({{b}^{2}}-{{a}^{2}})\,\sin \,2x}\]
Correct Answer: B
Solution :
If\[\int{f(x).\sin x.\cos x\,dx}\] \[=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log f(x)+c\] LHS \[\frac{1}{2}\int{f(x).(2\sin x\cos x)}dx\] \[=\frac{1}{2}\int{f(x).\sin 2x.dx}\] \[\left[ Here,put(x)=\frac{2}{({{b}^{2}}-{{a}^{2}})}\times \frac{1}{\cos 2x} \right]\] \[=\frac{1}{2}\int{\frac{2}{({{b}^{2}}-{{a}^{2}})}.\frac{\sin x}{\cos 2x}}dx\] \[=\frac{1}{({{b}^{2}}-{{a}^{2}})}\int{\tan 2x}dx\] \[=\frac{1}{({{b}^{2}}-{{a}^{2}})}.\frac{\log \sec 2x}{2}+{{c}_{1}}\] \[=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log \left( \frac{1}{\cos 2x} \right)+{{c}_{1}}\] \[\left[ put{{c}_{1}}=c+\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log \frac{2}{({{b}^{2}}-{{a}^{2}})} \right]\] \[=\frac{1}{2({{b}^{2}}-{{a}^{2}})}.\log \left( \frac{1}{\cos x} \right)\] \[+\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log \left( \frac{2}{({{b}^{2}}-{{a}^{2}})} \right)+c\] \[=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log \left[ \frac{2}{({{b}^{2}}-{{a}^{2}})\cos 2x} \right]+c\] Hence, \[f(x)=\frac{2}{({{b}^{2}}-{{a}^{2}})\cos 2x}\]
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