If _ ^ f(x) x x ;dx = 1 2( b^2 - a^2 ) (f(x)) + c, then f(x) = - Vidyadip

MCQ

If $\int_{}^{} {f(x)\sin x\cos x\;dx = \frac{1}{{2({b^2} - {a^2})}}\log (f(x))} + c$, then $f(x) = $

✓
$\frac{1}{{{a^2}{{\sin }^2}x + {b^2}{{\cos }^2}x}}$
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
$\frac{1}{{{a^2}{{\cos }^2}x - {b^2}{{\sin }^2}x}}$

✓

Answer

Correct option: A.

$\frac{1}{{{a^2}{{\sin }^2}x + {b^2}{{\cos }^2}x}}$

a
(a) Since $\int_{}^{} {f(x)\,\sin x\cos x\,dx} = \frac{1}{{2({b^2} - {a^2})}}\log \left( {f(x)} \right) + c$
Therefore $f(x)\sin x\cos x\, = \frac{1}{{2({b^2} - {a^2})}}.\frac{1}{{f(x)}}f'(x)$
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_{}^{} {(2{b^2}\sin x\cos x - 2{a^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)}}$.

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