Question

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

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

Answer 1

Answer: (C)

$-\cos 2x+c$

Let, $f\left(x\right)=\underset{y\to x}{lim}\frac{{\sin }^{2}y-{\sin }^{2}x}{y^2-x^2}$ $\left(\frac{0}{0}\right)$ form.

Now applying L'Hospital's rule we get,

$f\left(x\right)=\underset{y\to x}{lim}\frac{2\sin y.\cos y}{2y}$

or, $f\left(x\right)=\frac{\sin 2x}{2x}$........(1).

Now

$\int 4xf\left(x\right)dx$

$=$$\int 2\sin 2xdx$

$=-\cos 2x+c$. [Where $c$ being integrating constant]