Question

If $f\left(x\right)sinxcosxdx=\frac{1}{2(b^2-a^2)}log\left(f\right(x\left)\right)+c$, then f(x) =

• A. $\frac{1}{a^{2}si{n}^{2}x+b^{2}co{s}^{2}x}$
• B. $\frac{1}{a^{2}si{n}^{2}x-b^{2}co{s}^{2}x}$
• C. $\frac{1}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}$
• D. $\frac{1}{a^{2}co{s}^{2}x-b^{2}si{n}^{2}x}$

Answer 1

The correct option is A $\frac{1}{a^{2}si{n}^{2}x+b^{2}co{s}^{2}x}$

Since $\int f\left(x\right)sinxcosxdx=\frac{1}{2(b^2-a^2)}log\left(f\right(x\left)\right)+c$
Therefore $f\left(x\right)sinxcosx=\frac{1}{2(b^2-a^2)}.\frac{1}{f\left(x\right)}{f}^{{}^{\prime }}\left(x\right)$
Differentiating both sides w.r.t x
$\Rightarrow 2(b^2-a^2)sinxcosx=\frac{f^{{}^{\prime }}\left(x\right)}{f(x{)}^2}$
$\Rightarrow \int (2b^{2}sinxcosx-2a^{2}sinxcosx)dx=\int \frac{f^{{}^{\prime }}\left(x\right)}{\left\{f\right(x{)}^2\}}dx$
$\Rightarrow (-b^{2}co{s}^{2}x-a^{2}si{n}^{2}x)=-\frac{1}{f\left(x\right)}$
$\Rightarrow f\left(x\right)=\frac{1}{(a^{2}si{n}^{2}x+b^{2}co{s}^{2}x)}$