Question

Find $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{0}}^{{\mathbf{x}}^{\mathbf{2}}}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\mathbf{t}}{\mathbf{t}}\mathit{d}\mathit{t}$.

Short answer

The value of$\frac{d}{dx}{\int }_0^{x^2}\frac{\sin xt}{t}dt\text{is}\frac{3}{x}\sin x^3$ .

Step-by-step solution

Given Information

Given that . $\frac{d}{dx}{\int }_0^{x^2}\frac{\sin xt}{t}dt$ ...(1)

Formula Used

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

Solving the given equation

Compare equation (1) and equation (2).

$$\begin{gathered} \frac{d}{dx}{\int }_0^{x^2}\frac{\sin xt}{t}dt=\frac{\sin \left(x.x^2\right)}{x^2}.\frac{d}{dx}\left(x^2\right)-\frac{\sin \left(x.0\right)}{0}.\frac{d}{dx}\left(0\right)+{\int }_0^{x^2}t\frac{\cos xt}{t}dt \\ \frac{d}{dx}{\int }_0^{x^2}\frac{\sin xt}{t}dt=\frac{\sin x^3}{x^2}.\left(2x\right)-\frac{\sin 0}{0}.\left(0\right)+{\int }_0^{x^2}\cos xtdt \\ \frac{d}{dx}{\int }_0^{x^2}\frac{\sin xt}{t}dt=2x\frac{\sin x^3}{x^2}+{\left(\frac{\sin xt}{x}\right)}_0^{x^2} \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{d}{dx}{\int }_0^{x^2}\frac{\sin xt}{t}dt=2x\frac{\sin x^3}{x^2}+\frac{\sin \left(x.x^2\right)}{x}-\frac{\sin \left(x.0\right)}{x} \\ \frac{d}{dx}{\int }_0^{x^2}\frac{\sin xt}{t}dt=\frac{2\sin x^3}{x}+\frac{\sin x^3}{x} \\ \frac{d}{dx}{\int }_0^{x^2}\frac{\sin xt}{t}dt=\frac{2}{x}\sin x^3+\frac{1}{x}\sin x^3 \\ \frac{d}{dx}{\int }_0^{x^2}\frac{\sin xt}{t}dt=\frac{3}{x}\sin x^3 \end{gathered}$$

Hence the value of $\frac{d}{dx}{\int }_0^{x^2}\frac{\sin xt}{t}dt=\frac{3}{x}\sin x^3$.