Question

Combining derivatives and integrals: Simplify each of the following as much as possible:

$\frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt$

Short answer

The solution is$\frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt=\frac{3}{4x}\left(\sin \right(\ln x)-\frac{1}{3}\sin 3(\ln x\left)\right).$

Step-by-step solution

Step 1. Given information

Integral:$\frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt$

Step 2. Calculation

The given equation is-

$\frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt$

Using ${\sin }^{3}x=\frac{3}{4}\sin x-\frac{1}{4}\sin 3x$

$$\begin{gathered} \frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt=\frac{d}{dx}{\int }_0^{\ln x}(\frac{3}{4}\sin t-\frac{1}{4}\sin 3t)dt \\ \frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt=\frac{d}{dx}\left(\frac{3}{4}{\int }_0^{\ln x}\right(\sin t)dt-\frac{1}{4}{\int }_0^{\ln x}(\sin 3t\left)dt\right) \\ U\sin g\int \sin xdx=-\cos x+C \\ \frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt=\frac{d}{dx}(\frac{3}{4}{[-\cos t]}_0^{\ln x}-\frac{1}{4}{[-\frac{1}{3}\cos 3t]}_0^{\ln x})\frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt=\frac{d}{dx}(-\frac{3}{4}[\cos (\ln x)-\cos 0]+\frac{1}{12}[\cos 3(\ln x)-\cos 0\left]\right) \\ \frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt=(-\frac{3}{4}\frac{d}{dx}[\cos (\ln x)-1]+\frac{1}{12}\frac{d}{dx}[\cos 3(\ln x)-1\left]\right) \\ U\sin g\frac{d}{dx}\cos x=-\sin x \\ \frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt=(-\frac{3}{4}[-\sin (\ln x)\frac{d}{dx}\ln x-0]+\frac{1}{12}[-\sin 3(\ln x)\frac{d}{dx}3\ln x\left]\right) \\ \frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt=\left(\frac{3}{4}\right[\frac{\sin (\ln x)}{x}]-\frac{3}{12}[\frac{\sin (\ln x)}{x}\left]\right) \\ \frac{d}{dx}{\int }_0^{\ln x}{\sin }^{3}tdt=\frac{3}{4x}\left(\sin \right(\ln x)-\frac{1}{3}\sin 3(\ln x\left)\right) \end{gathered}$$