Question

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

$\frac{d}{dx}{\int }_x^3{\sin }^{3}tdt$

Short answer

The solution is$\frac{d}{dx}{\int }_x^3{\sin }^{3}tdt=(-\frac{3}{4}[\sin x]+\frac{1}{4}[\sin 3x\left]\right)$

Step-by-step solution

Step 1. Given information

Integral:$\frac{d}{dx}{\int }_x^3{\sin }^{3}tdt$

Step 2. Calculation

The given integration is $\frac{d}{dx}{\int }_x^3{\sin }^{3}tdt$

$\frac{d}{dx}{\int }_x^3{\sin }^{3}tdt$

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

$$\begin{gathered} =\frac{d}{dx}{\int }_x^3(\frac{3}{4}\sin t-\frac{1}{4}\sin 3t)dt \\ =\frac{d}{dx}\int \left(\frac{3}{4}{\int }_x^3\right(\sin t)dt-\frac{1}{4}{\int }_x^3(\sin 3t\left)dt\right) \\ U\sin g\int \sin xdx=-\cos x+C \\ =\frac{d}{dx}(\frac{3}{4}{[-\cos t]}_x^3-\frac{1}{4}{[-\frac{1}{3}\cos 3t]}_x^3)=\frac{d}{dx}(-\frac{3}{4}[\cos 3-\cos x]+\frac{1}{12}[\cos 9-\cos 3x\left]\right) \\ =(-\frac{3}{4}\frac{d}{dx}[\cos 3-\cos x]+\frac{1}{12}\frac{d}{dx}[\cos 9-\cos 3x] \\ u\sin g\frac{d}{dx}\cos x=-\sin x \\ =(-\frac{3}{4}\left[\sin x\right]+\frac{1}{12}\left[3\sin 3x\right]) \\ =(\frac{3}{4}\left[sinx\right]+\frac{1}{4}\left[\sin 3x\right]) \end{gathered}$$