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Question: If \[f(x) = {\lim _{y \to x}}\dfrac{{[{{\sin }^2}y - {{\sin }^2}x]}}{{[{y^2} - {x^2}]}}\], then \[\i...

If f(x)=lim⁡y→x[sin⁡2y−sin⁡2x][y2−x2]f(x) = {\lim _{y \to x}}\dfrac{{[{{\sin }^2}y - {{\sin }^2}x]}}{{[{y^2} - {x^2}]}}, then ∫4xf(x)dx=\int {4xf(x)} dx =
A. cos⁡(2x)+c\cos (2x) + c
B. 2cos⁡(2x)+c2\cos (2x) + c
C. −cos⁡(2x)+c - \cos (2x) + c
D. −2cos⁡(2x)+c - 2\cos (2x) + c

Explanation

Solution

This question involves the concepts of limits and integration. Firstly, we need to find out the value of the function using limit and then replace it in the integration symbol. The property of limit used are:
f(x)=lim⁡x→0sin⁡(x)xf(x) = {\lim _{x \to 0}}\dfrac{{\sin (x)}}{x},
lim⁡x→h(x)f(x)+g(x)=lim⁡x→h(x)f(x)+lim⁡x→h(x)g(x){\lim _{x \to h(x)}}f(x) + g(x) = {\lim _{x \to h(x)}}f(x) + {\lim _{x \to h(x)}}g(x)
Where the limit of these functions exists independently.

Complete step by step answer:
Now, let us begin the question by solving the value of the limit
⇒f(x)=lim⁡y→x[sin⁡2y−sin⁡2x][y2−x2],\Rightarrow f(x) = {\lim _{y \to x}}\dfrac{{[{{\sin }^2}y - {{\sin }^2}x]}}{{[{y^2} - {x^2}]}},
Now, by applying the property on both the numerator and the denominator we get,
⇒f(x)=lim⁡y→x[sin⁡y−sin⁡x]×[sin⁡y+sin⁡x][y−x]×[y+x]\Rightarrow f(x) = {\lim _{y \to x}}\dfrac{{[\sin y - \sin x] \times [\sin y + \sin x]}}{{[y - x] \times [y + x]}}
Now, by applying the trigonometric identity sin⁡y−sin⁡x=2[sin⁡(y−x2)cos⁡(y+x2)]\sin y - \sin x = 2[\sin (\dfrac{{y - x}}{2})\cos (\dfrac{{y + x}}{2})] and sin⁡y+sin⁡x=2[sin⁡(y+x2)cos⁡(y−x2)]\sin y + \sin x = 2[\sin (\dfrac{{y + x}}{2})\cos (\dfrac{{y - x}}{2})]we get,
⇒f(x)=lim⁡y→x2[sin⁡(y−x2)cos⁡(y+x2)]×2[sin⁡(y+x2)cos⁡(y−x2)][y−x]×[y+x]\Rightarrow f(x) = {\lim _{y \to x}}\dfrac{{2[\sin (\dfrac{{y - x}}{2})\cos (\dfrac{{y + x}}{2})] \times 2[\sin (\dfrac{{y + x}}{2})\cos (\dfrac{{y - x}}{2})]}}{{[y - x] \times [y + x]}}

Now, by grouping sin⁡(y−x2)\sin (\dfrac{{y - x}}{2}) with cos⁡(y−x2)\cos (\dfrac{{y - x}}{2}) and sin⁡(y+x2)\sin (\dfrac{{y + x}}{2})withcos⁡(y+x2)\cos (\dfrac{{y + x}}{2}) we get,
⇒f(x)=lim⁡y→x2[sin⁡(y−x2)cos⁡(y−x2)]×2[sin⁡(y+x2)cos⁡(y+x2)][y−x]×[y+x]\Rightarrow f(x) = {\lim _{y \to x}}\dfrac{{2[\sin (\dfrac{{y - x}}{2})\cos (\dfrac{{y - x}}{2})] \times 2[\sin (\dfrac{{y + x}}{2})\cos (\dfrac{{y + x}}{2})]}}{{[y - x] \times [y + x]}}
Now, by applying the trigonometric property, sin⁡(2x)=sin⁡(x)cos⁡(x)\sin (2x) = \sin (x)\cos (x) we get,
⇒f(x)=lim⁡y→x[sin⁡(y−x)]×[sin⁡(y+x)][y−x]×[y+x]\Rightarrow f(x) = {\lim _{y \to x}}\dfrac{{[\sin (y - x)] \times [\sin (y + x)]}}{{[y - x] \times [y + x]}}
Now, by separating the two fractions as their limit exists independently, we get,
⇒f(x)=lim⁡y→x[sin⁡(y−x)][y−x]×lim⁡y→x[sin⁡(y+x)][y+x]\Rightarrow f(x) = {\lim _{y \to x}}\dfrac{{[\sin (y - x)]}}{{[y - x]}} \times {\lim _{y \to x}}\dfrac{{[\sin (y + x)]}}{{[y + x]}}

And finally, we get the value of the function,
⇒f(x)=sin⁡(2x)2x\Rightarrow f(x) = \dfrac{{\sin (2x)}}{{2x}}
Now, we have successfully calculated the value of the function.Let us move forward to the next step of the problem. Now, by replacing this value of the function in the integral, we get,
⇒∫4xf(x)dx=∫4xsin⁡(2x)2xdx\Rightarrow \int {4xf(x)} dx = \int {4x\dfrac{{\sin (2x)}}{{2x}}} dx
Now, by eliminating the parameter x, we get,
⇒∫2sin⁡(2x)dx\Rightarrow \int {2\sin (2x)dx}
Now, integrating the function we get,
⇒−2cos⁡(2x)2+c\Rightarrow - 2\dfrac{{\cos (2x)}}{2} + c
Now, by eliminating two from both the numerator and the denominator we get,
∴−cos⁡(2x)+c\therefore - \cos (2x) + c Where cc is an arbitrary constant.

Therefore, option C is the correct answer.

Note: This question is a little concerning concepts from multiple topics like trigonometry, limits, and integration. One should be well versed with the theory and the application of these topics to solve this question. Do not commit calculation mistakes, and be sure of the final answer.