Problem of the Week

Updated at Dec 23, 2013 12:50 PM

This week's problem comes from the calculus category.

How can we solve for the integral of cos⁡3x\cos^{3}x?

Let's begin!



∫cos⁡3x dx\int \cos^{3}x \, dx

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

2
Use Integration by Substitution.
Let u=sin⁡xu=\sin{x}, du=cos⁡x dxdu=\cos{x} \, dx

3
Using uu and dudu above, rewrite ∫(1−sin⁡2x)cos⁡x dx\int (1-\sin^{2}x)\cos{x} \, dx.
∫1−u2 du\int 1-{u}^{2} \, du

4
Use Power Rule: ∫xn dx=xn+1n+1+C\int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C.
u−u33u-\frac{{u}^{3}}{3}

5
Substitute u=sin⁡xu=\sin{x} back into the original integral.
sin⁡x−sin⁡3x3\sin{x}-\frac{\sin^{3}x}{3}

6
Add constant.
sin⁡x−sin⁡3x3+C\sin{x}-\frac{\sin^{3}x}{3}+C

Done
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