Question

If $\int f\left(x\right)\sin x\cos xdx=\frac{1}{2(b^2-a^2)}\log f\left(x\right)+c$, where $c$ is the constant of integration,

then $f\left(x\right)=?$

• A. $\frac{2}{(b^2-a^2)\sin 2x}$
• B. $\frac{2}{ab\sin 2x}$
• C. $\frac{2}{(b^2-a^2)\cos 2x}$
• D. $\frac{2}{ab\cos 2x}$

Answer 1

The correct option is C $\frac{2}{(b^2-a^2)\cos 2x}$

$\int f\left(x\right)sinxcosxdx=\frac{1}{2(b^2-a^2)}lnf\left(x\right)+c$

Differentiating above equation we get

$f\left(x\right)sinxcosx=\frac{1}{2(b^2-a^2)}\frac{1}{f\left(x\right)}{f}^{{}^{\prime }}\left(x\right)$

$⟹\frac{f^{{}^{\prime }}x}{f(x{)}^2}=2(b^2-a^2)sinxcosx$

$⟹\frac{f^{{}^{\prime }}x}{f(x{)}^2}=(b^2-a^2)sin2x$

Integrating above equation with respect to x we get

$\int \frac{f^{{}^{\prime }}\left(x\right)}{f(x{)}^2}dx=\int (b^2-a^2)sin2xdx$

$⟹-\frac{1}{f\left(x\right)}=-(b^2-a^2)\frac{cos2x}{2}$

$f\left(x\right)=\frac{2}{(b^2-a^2)cos2x}$

Hence, the answer is option (C)/.