Chapter 4: Q. 46 (page 405)

Combining derivatives and integrals: Simplify each of the following as much as possible:
$\frac{d}{d x} \int_{x}^{3} \sin^{3} t d t$

Short Answer

Expert verified

The solution is $\frac{d}{d x} \int_{x}^{3} \sin^{3} t d t = (- \frac{3}{4} [\sin x] + \frac{1}{4} [\sin 3 x])$

Step by step solution

Step 1. Given information

Integral: $\frac{d}{d x} \int_{x}^{3} \sin^{3} t d t$

Step 2. Calculation

The given integration is $\frac{d}{d x} \int_{x}^{3} \sin^{3} t d t$

$\frac{d}{d x} \int_{x}^{3} \sin^{3} t d t$

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

$= \frac{d}{d x} \int_{x}^{3} (\frac{3}{4} \sin t - \frac{1}{4} \sin 3 t) d t = \frac{d}{d x} \int (\frac{3}{4} \int_{x}^{3} (\sin t) d t - \frac{1}{4} \int_{x}^{3} (\sin 3 t) d t) U \sin g \int \sin x d x = - \cos x + C = \frac{d}{d x} (\frac{3}{4} {[- \cos t]}_{x}^{3} - \frac{1}{4} {[- \frac{1}{3} \cos 3 t]}_{x}^{3}) = \frac{d}{d x} (- \frac{3}{4} [\cos 3 - \cos x] + \frac{1}{12} [\cos 9 - \cos 3 x]) = (- \frac{3}{4} \frac{d}{d x} [\cos 3 - \cos x] + \frac{1}{12} \frac{d}{d x} [\cos 9 - \cos 3 x] u \sin g \frac{d}{d x} \cos x = - \sin x = (- \frac{3}{4} [\sin x] + \frac{1}{12} [3 \sin 3 x]) = (\frac{3}{4} [s i n x] + \frac{1}{4} [\sin 3 x])$

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Combining derivatives and integrals: Simplify each of the following as much as possible:
$\frac{d}{d x} \int_{x}^{3} \sin^{3} t d t$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

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Combining derivatives and integrals: Simplify each of the following as much as possible:ddx∫x3sin3tdt

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation

One App. One Place for Learning.

Most popular questions from this chapter

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Combining derivatives and integrals: Simplify each of the following as much as possible:
$\frac{d}{d x} \int_{x}^{3} \sin^{3} t d t$