Question

If $f\left(x\right)=li{m}_{y\to x}\frac{{\sin }^{2}y-{\sin }^{2}x}{y^2-x^2}$, then $\int 4xf\left(x\right)dx=$

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

Answer 1

The correct option is C

$–\cos 2x+C$

Explanation for correct option:

Find the value of $\int 4xf\left(x\right)dx$:

Given expression,

$f\left(x\right)=li{m}_{y\to x}\frac{{\sin }^{2}y-{\sin }^{2}x}{y^2-x^2}$

Since, the function form $\left(\frac{0}{0}\right)$ at $y=x$, then

Applying L' Hospital rule,

$\begin{array}{rcl}f\left(x\right)& =& li{m}_{y\to x}\frac{2\sin y\cos y}{2y-0}\\ & =& li{m}_{y\to x}\frac{2\sin y}{2y}\cos y\\ & =& li{m}_{y\to x}\frac{\sin 2y}{2y}\\ & =& \frac{\sin 2x}{2x}\end{array}$

Now,

$\begin{array}{rcl}\int 4xf\left(x\right)dx& =& \int \frac{4x\sin 2x}{2x}dx\\ & =& 2\int \sin 2xdx\\ & =& -\frac{2\cos 2x}{2}+C\\ & =& -\cos 2x+C\end{array}$

Hence, the correct option is C.