Integral Calculus Question 329
Question: If $ \int_{{}}^{{}}{f(x)\sin x\cos x\ dx=\frac{1}{2(b^{2}-a^{2})}\log (f(x))}+c $ , then $ f(x)= $
Options:
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} $
Show Answer
Answer:
Correct Answer: A
Solution:
Since $ \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) $
Differentiating both sides $ w.r.t.x $
$ \Rightarrow 2(b^{2}-a^{2})\sin x\cos x=\frac{{f}’(x)}{f{{(x)}^{2}}} $
$ \Rightarrow \int_{{}}^{{}}{(2b^{2}\sin x\cos x-2a^{2}\sin x\cos x),dx=}\int_{{}}^{{}}{\frac{{f}’(x)}{{{{f(x)}}^{2}}}},dx $
$ \Rightarrow \pm ,(-b^{2}{{\cos }^{2}}x-a^{2}{{\sin }^{2}}x)=-\frac{1}{f(x)} $
$ \Rightarrow f(x)=\pm \frac{1}{(a^{2}{{\sin }^{2}}x+b^{2}{{\cos }^{2}}x)} $ .
BETA
