Question

Find $\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{t}}\frac{\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{x}}{\mathbf{x}}\mathit{d}\mathit{x}$.

Short answer

The value of $\frac{d}{dt}{\int }_0^{\sin t}\frac{{\sin }^{-1}x}{x}dx\text{is}t\cot t$.

Step-by-step solution

Given Information

Given that .$\frac{d}{dt}{\int }_0^{\sin t}\frac{{\sin }^{-1}x}{x}dx$ ...(1)

Formula Used

We know that$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{u}\left(x\right)}^{\mathbf{v}\left(x\right)}\mathit{f}\left(x,t\right)\mathit{d}\mathit{t}\mathbf{=}\mathit{f}\left(x,v\right)\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathit{f}\left(x,u\right)\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}{\mathbf{\int }}_{\mathbf{u}}^{\mathbf{v}}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{x}}\mathit{d}\mathit{t}\mathbf{\in }$ ...(2)

Solve the given equation

Compare equation (1) and equation (2).

$\frac{d}{dt}{\int }_0^{\sin t}\frac{{\sin }^{-1}x}{x}dx=\frac{{\sin }^{-1}\left(\sin t\right)}{\sin t}.\frac{d}{dt}\left(\sin t\right)-\frac{{\sin }^{-1}\left(0\right)}{0}.\frac{d}{dt}\left(0\right)+{\int }_0^{\sin t}0dx$

Therefore,

$$\begin{gathered} \frac{d}{dt}{\int }_0^{\sin t}\frac{{\sin }^{-1}x}{x}dx=\frac{t}{\sin t}.\left(\cos t\right) \\ \frac{d}{dt}{\int }_0^{\sin t}\frac{{\sin }^{-1}x}{x}dx=t\left(\frac{\cos t}{\sin t}\right) \\ \frac{d}{dt}{\int }_0^{\sin t}\frac{{\sin }^{-1}x}{x}dx=t\cot t \end{gathered}$$

Hence the value of$\frac{d}{dt}{\int }_0^{\sin t}\frac{{\sin }^{-1}x}{x}dx=t\cot t$ .