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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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:

\begin{align*} \int\csc^2x\;dx&=-\csc x\cot x-\int\csc^3x\;dx-\ln|\csc x+\cot x|+C \end{align*}

Add $\displaystyle\int\csc^3x\;dx$ on both sides and then divide both sides by 2:

\begin{align*} 2\int\csc^3x\;dx&=-\csc x\cot x-\ln|\csc x+\cot x|+C\\ \int\csc^3x\;dx&=-\dfrac{1}{2}\csc x\cot x-\dfrac{1}{2}\ln|\csc x+\cot x|+C \end{align*}

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