Related questions with answers

The alternation of dark and light wood produces what we commonly call $___________$

A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of $6 \times 10^{-8}\ \mathrm{C}$ at location $\langle- 0.3,0,0\rangle\ \mathrm{m} .$ At location $\langle 0,0.07,0\rangle\ \mathrm{m},$ what is the electric field contributed by the polarization charges on the surface of the metal sphere? How do you know?

The stockholders' equity of Palepu Company at December 31. 2018. appears below.

Common stock, $10 par value, 200,000 shares authorized;
80,000 shares issued and outstanding	$800,000
Paid-in capital in excess of par value	480,000
Retained earnings	305,000

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May 12 Declared and issued a 7% stock dividend; the common stock market value was $18 per share.

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Using the financial statement effects template, illustrate the effects of these transactions.

Question

Use integration by parts to find $\int \csc ^{3} x d x$ . (Hint: Start with $\int \csc ^{3} x d x=\int \csc x \cdot \csc ^{2} x d x$ .)

Solution

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Step 1

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Rewrite $\csc^3x$ using the hint:

\int\csc^3x\;dx=\int\csc x\cdot\csc^2x\;dx

Use the integration by parts formula $\displaystyle\int u\;dv=uv-\int v\;du$ with $u=\csc x$ and $dv=\csc^2x\;dx$ . Then $du=-\csc x\cot x\;dx$ and $v=-\cot x$ :

\begin{align*} \int\csc^3x\;dx&=\csc x(-\cot x)-\int(-\cot x)(-\csc x\cot x\;dx)\\ &=-\csc x\cot x-\int\csc x\cot^2x\;dx \end{align*}

Use the trig identity $1+\cot^2x=\csc^2x$ to replace $\cot^2x$ with $\csc^2x-1$ . Then distribute:

\begin{align*} \int\csc^3x\;dx&=-\csc x\cot x-\int\csc x(\csc^2x-1)\;dx\\ &=-\csc x\cot x-\int(\csc^3x-\csc x)\;dx\\ &=-\csc x\cot x-\int\csc^3x\;dx+\int\csc x\;dx \end{align*}

Integrate $\displaystyle\int\csc x\;dx$ using the rule $\displaystyle\int\csc x\;dx=-\ln|\csc x+\cot x|+C$ that you proved in Exercise 46 of the previous section: