Thread: Integral help

Trying to integrate the following

([ will stand for the integral sign)

[|cos(x)|^3 dx between 0 and 2pi


Now, Maple tells me the answer should be 8/3, but I keep getting 0.

[|cos(x)|^3 dx = [cos^2(x).cos(x) dx

= [(1-sin^2(x)].cos(x)dx

= [cos(x)-cos(x)sin^2(x) dx

= sin(x) -sin^3/3

which gives me 0 when I evaluate it. Any help would really be appreciated. Thank you

You're ignoring the absolute value signs in $\displaystyle |\cos^3x|$. In any interval of length 2π, the function $\displaystyle \cos^3x$ is positive through half the interval and negative through the other half, so its integral over the whole interval is 0. But if you take its absolute value, that will always be positive. So the value of the integral will be twice the value of the integral of $\displaystyle \cos^3x$ over the half of the interval in which it is positive (sketch a graph if this helps you to visualise it).

The function $\displaystyle \cos^3x$ is positive in the interval $\displaystyle -\pi/2\leqslant x\leqslant\pi/2$, so use your expression for the indefinite integral (which by the way is a very good method of doing it), evaluate it between $\displaystyle -\pi/2$ and $\displaystyle \pi/2$, and multiply the answer by 2.

Originally Posted by Ratchet1010

Trying to integrate the following

([ will stand for the integral sign)

[|cos(x)|^3 dx between 0 and 2pi


Now, Maple tells me the answer should be 8/3, but I keep getting 0.

[|cos(x)|^3 dx = [cos^2(x).cos(x) dx

= [(1-sin^2(x)].cos(x)dx

= [cos(x)-cos(x)sin^2(x) dx

= sin(x) -sin^3/3

which gives me 0 when I evaluate it. Any help would really be appreciated. Thank you

What you were calculating was net area. What you want to do is calculate total area.

Note that $\displaystyle \left|\cos x\right|=\left\{\begin{array}{rl}\cos x & \text{when }0\leq x\leq\tfrac{\pi}{2}\,\&\,\tfrac{3\pi}{2}\leq x\leq 2\pi\\-\cos x & \text{when }\tfrac{\pi}{2}\leq x\leq \tfrac{3\pi}{2}\end{array}\right.$

So it follows that $\displaystyle \int_0^{2\pi}\left|\cos x\right|^3\,dx=\int_0^{\frac{\pi}{2}}\cos^3x\,dx-\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\cos^3x\,dx+\ int_{\frac{3\pi}{2}}^{2\pi}\cos^3x\,dx=4\int_{0}^{ \frac{\pi}{2}}\cos^3x\,dx$

Can you take it from here?

You can reuse the work you already did, but just use the limits of integration I came up with.