If \[f:( - 1,1) \to R\] be a continuous function. If \[\int\limits_0^{\sin x} {f(t).dt} = \dfrac{{\sqrt 3 }}{2}x\] ,then \[f(\dfrac{{\sqrt 3 }}{2})\] is equal to
A. \[\dfrac{{\sqrt 3 }}{2}\]
B. \[\sqrt 3 \]
C. \[\dfrac{1}{2}\]
D. \[\sqrt {\dfrac{3}{2}} \]

Hint: From the concept of definite integral calculus, we can apply the formula of Leibnitz’s integral calculation as
 \[\int\limits_{f(x)}^{g(x)} {z(t).dt} = z(g(x))g'(x) - z(f(x))f'(x)\] . So, for the above given equation differentiate both sides with respect to x. And then use proper substitution in order to get the required term and hence solve as per calculation below.

Complete step-by-step answer:
As the given equation is as \[\int\limits_0^{\sin x} {f(t).dt} = \dfrac{{\sqrt 3 }}{2}x\]
Now, differentiate both sides with respect to x.
So, it can be simplified as,
Firstly, differentiate L.H.S as,
 \[
   \Rightarrow \int\limits_0^{\sin x} {f(t).dt} = f(\sin x)\cos x - f(0)\dfrac{{d0}}{{dx}} \\
   \Rightarrow \int\limits_0^{\sin x} {f(t).dt} = f(\sin x)\cos x \\
 \]
 Now, differentiation of R.H.S is
  \[ \Rightarrow \dfrac{d}{{dx}}(\dfrac{{\sqrt 3 x}}{2}) = \dfrac{{\sqrt 3 }}{2}\]
So, the equation can be formed as,
 \[ \Rightarrow f(\sin x)\cos x = \dfrac{{\sqrt 3 }}{2}\]
Now, as we have to find the value of \[f(\dfrac{{\sqrt 3 }}{2})\] so on comparison with the above equation we can see that it is possible only when we take, \[{\text{x = }}\dfrac{{{\pi }}}{{\text{3}}}\] . So, substituting the value of x in the above equation.
 \[
   \Rightarrow f(\sin x)\cos x = \dfrac{{\sqrt 3 }}{2} \\
   \Rightarrow f(\sin (\dfrac{\pi }{3}))\cos (\dfrac{\pi }{3}) = \dfrac{{\sqrt 3 }}{2} \\
 \]
As, \[\sin (\dfrac{\pi }{3}) = \dfrac{{\sqrt 3 }}{2},\cos (\dfrac{\pi }{3}) = \dfrac{1}{2}\] , we get,
 \[ \Rightarrow f{\text{(}}\dfrac{{\sqrt {\text{3}} }}{{\text{2}}}{\text{)}}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = }}\dfrac{{\sqrt {\text{3}} }}{{\text{2}}}\]
On simplifying further, we get,
 \[ \Rightarrow f(\dfrac{{\sqrt 3 }}{2}) = \sqrt 3 \]
Hence, option (B) is our required answer.


Note: The Leibniz integral rule gives a formula for differentiation of a definite integral whose limits are functions of the differential variable, (1) It is sometimes known as differentiation under the integral sign. This rule can be used to evaluate certain unusual definite integrals such as \[\int\limits_{f(x)}^{g(x)} {z(t).dt} = z(g(x))g'(x) - z(f(x))f'(x)\]. Thus, do the proper differentiation and remember trigonometric values in order to avoid mistakes.
Also we need to know that,
Leibniz rule for double integral variable can be given as
\[\dfrac{d}{{dx}}\int\limits_{\alpha (x)}^{\beta (x)} {f(x,y)} dy = f(x,\beta (x))\beta '(x) - f(x,\alpha (x))\alpha '(x) + \int\limits_{\alpha (x)}^{\beta (x)} {\dfrac{{\delta f}}{{\delta x}}} dy\]
Thus, in the following way we can proceed for the above given question.