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 :
| \[\int{f(x)\,\sin x\,\cos x\,dx=\frac{1}{2({{b}^{2}}-{{a}^{2}})}\log \,\left( f(x) \right)+C}\] |
| 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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