Question

If $f(x)=\underset{y\to x}{\lim }\frac{{\sin }^{2}y-{\sin }^{2}x}{y^2-x^2},$ then $\int 4xf(x)dx$ =

• A. $\cos 2x+c$
• B. $2\cos 2x+c$
• C. $-\cos 2x+c$
• D. $-2\cos 2x+c$

Answer 1

Answer: (C)

$f(x)=\underset{y\to x}{\lim }\frac{{\sin }^{2}y-{\sin }^{2}x}{y^2-x^2}$
$\left[\frac{0}{0}form\right]$
$=\underset{y\to x}{\lim }\frac{2\sin y\cos y-0}{2y-0}$
$=\frac{\sin 2x}{2x}$
$\therefore$ $\int 4xf(x)dx=\int 4x\left(\frac{\sin 2x}{2x}\right)dx$
$=2\int \sin 2xdx$
$=-\cos 2x+c$