Problem of the Week

Updated at Oct 7, 2013 8:24 AM

To get more practice in calculus, we brought you this problem of the week:

How can we find the integral of \(\csc^{3}x\)?

Check out the solution below!



\[\int \csc^{3}x \, dx\]

1
Use Integration by Parts on \(\int \csc^{3}x \, dx\).
Let \(u=\csc{x}\), \(dv=\csc^{2}x\), \(du=-\csc{x}\cot{x} \, dx\), \(v=-\cot{x}\)

2
Substitute the above into \(uv-\int v \, du\).
\[-\csc{x}\cot{x}-\int \cot^{2}x\csc{x} \, dx\]

3
Use Pythagorean Identities: \(\cot^{2}x=\csc^{2}x-1\).
\[-\csc{x}\cot{x}-\int (\csc^{2}x-1)\csc{x} \, dx\]

4
Expand.
\[-\csc{x}\cot{x}-\int \csc^{3}x-\csc{x} \, dx\]

5
Use Sum Rule: \(\int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx\).
\[-\csc{x}\cot{x}-\int \csc^{3}x \, dx+\int \csc{x} \, dx\]

6
Set it as equal to the original integral \(\int \csc^{3}x \, dx\).
\[\int \csc^{3}x \, dx=-\csc{x}\cot{x}-\int \csc^{3}x \, dx+\int \csc{x} \, dx\]

7
Add \(\int \csc^{3}x \, dx\) to both sides.
\[\int \csc^{3}x \, dx+\int \csc^{3}x \, dx=-\csc{x}\cot{x}+\int \csc{x} \, dx\]

8
Simplify  \(\int \csc^{3}x \, dx+\int \csc^{3}x \, dx\)  to  \(2\int \csc^{3}x \, dx\).
\[2\int \csc^{3}x \, dx=-\csc{x}\cot{x}+\int \csc{x} \, dx\]

9
Divide both sides by \(2\).
\[\int \csc^{3}x \, dx=\frac{-\csc{x}\cot{x}+\int \csc{x} \, dx}{2}\]

10
Original integral solved.
\[\frac{-\csc{x}\cot{x}+\int \csc{x} \, dx}{2}\]

11
Use Trigonometric Integration: the integral of \(\csc{x}\) is \(\ln{(\csc{x}-\cot{x})}\).
\[\frac{-\csc{x}\cot{x}+\ln{(\csc{x}-\cot{x})}}{2}\]

12
Add constant.
\[\frac{-\csc{x}\cot{x}+\ln{(\csc{x}-\cot{x})}}{2}+C\]

Done
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