Question

Show that $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}}^{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{t}}^{\mathbf{2}}}\mathit{d}\mathit{t}\mathbf{=}\mathbf{1}$.

Short answer

It is proved that $\frac{d}{dx}{\int }_{\cos x}^{\sin x}\sqrt{1-t^2}dt=1$.

Step-by-step solution

Given Information

Given that . $\frac{d}{dx}{\int }_{\cos x}^{\sin x}\sqrt{1-t^2}dt$ ...(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).

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

Hence Proved