Problem of the Week

Updated at Feb 10, 2014 3:33 PM

This week we have another calculus problem:

How can we find the integral of cot⁡3x\cot^{3}x?

Let's start!



∫cot⁡3x dx\int \cot^{3}x \, dx

1
Use Pythagorean Identities: cot⁡2x=csc⁡2x−1\cot^{2}x=\csc^{2}x-1.
∫(csc⁡2x−1)cot⁡x dx\int (\csc^{2}x-1)\cot{x} \, dx

2
Expand.
∫cot⁡xcsc⁡2x−cot⁡x dx\int \cot{x}\csc^{2}x-\cot{x} \, dx

3
Use Sum Rule: ∫f(x)+g(x) dx=∫f(x) dx+∫g(x) dx\int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx.
∫cot⁡xcsc⁡2x dx−∫cot⁡x dx\int \cot{x}\csc^{2}x \, dx-\int \cot{x} \, dx

4
Simplify the trigonometric functions.
∫cos⁡xsin⁡3x dx−∫cot⁡x dx\int \frac{\cos{x}}{\sin^{3}x} \, dx-\int \cot{x} \, dx

5
Use Integration by Substitution on ∫cos⁡xsin⁡3x dx\int \frac{\cos{x}}{\sin^{3}x} \, dx.
Let u=sin⁡xu=\sin{x}, du=cos⁡x dxdu=\cos{x} \, dx

6
Using uu and dudu above, rewrite ∫cos⁡xsin⁡3x dx\int \frac{\cos{x}}{\sin^{3}x} \, dx.
∫1u3 du\int \frac{1}{{u}^{3}} \, du

7
Use Power Rule: ∫xn dx=xn+1n+1+C\int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C.
−12u2-\frac{1}{2{u}^{2}}

8
Substitute u=sin⁡xu=\sin{x} back into the original integral.
−12sin⁡2x-\frac{1}{2\sin^{2}x}

9
Rewrite the integral with the completed substitution.
−12sin⁡2x−∫cot⁡x dx-\frac{1}{2\sin^{2}x}-\int \cot{x} \, dx

10
Use Trigonometric Integration: the integral of cot⁡x\cot{x} is ln⁡(sin⁡x)\ln{(\sin{x})}.
−12sin⁡2x−ln⁡(sin⁡x)-\frac{1}{2\sin^{2}x}-\ln{(\sin{x})}

11
Add constant.
−12sin⁡2x−ln⁡(sin⁡x)+C-\frac{1}{2\sin^{2}x}-\ln{(\sin{x})}+C

Done
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