Integral Calculus Question 132
If $ \int{f(x)\sin x\cos xdx=\frac{1}{2(b^{2}-a^{2})}{\log_{e}}(f(x))+A,} $ $ b\ne \pm a, $ then $ {f(x)}^{-1} $ is equal to
Options:
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}sin^{2}x+C $
D) $ a^{2}{{\cos }^{2}}x-b^{2}sin^{2}x+C $
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Answer:
Correct Answer: A
Solution:
$ \int{f(x)\sin x\cos xdx=\frac{1}{2}\log (f(x))+C} $ Therefore $ f(x)\sin x\cos x=\frac{1}{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)}{{{( f(x) )}^{2}}} $ $ \int{(2b^{2}\sin x\cos x-2a^{2}\sin x\cos x)dx=\int{\frac{f’(x)}{{{( f(x) )}^{2}}}dx}} $ [by integrating both the sides]
$ \Rightarrow -b^{2}{{\cos }^{2}}x-a^{2}{{\sin }^{2}}x-C=-\frac{1}{f(x)} $
BETA
