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I am having trouble changing variables and converting real trig integrals into the complex plane. I know these integrals need to be converted to contour integrals along the circle $|z|=1$ but I am having trouble changing the bound if the integrals are not already over the interval $[0,2 \pi]$.

For instance the book I am using, Saff Snider says if the given integral is: $$ \int_{0}^{\pi}\frac{8}{5+2 \cos \theta} d \theta$$

The book states since $\cos(\theta)=\cos (\theta -2 \pi)$ This implies $$ \int_{0}^{\pi}\frac{8}{5+2 \cos \theta} d \theta=\frac{1}{2}\int_{0}^{2\pi}\frac{8}{5+2 \cos \theta} d \theta$$

I know this has something to do with the evenness and periodicity of the cosine function. Can someone explain this to me?

Also another problem in the book is:

$$\int_{- \pi}^{\pi}\frac{1}{1+\sin^{2} \theta} d \theta$$ and this integral is equal to $$\int_{0}^{2\pi}\frac{1}{1+\sin^{2} \theta} d \theta$$

I have also seen the integral: $$\int_{0}^{\frac{\pi}{2}}\frac{\cos 2 \theta}{1+2\cos^{2} \theta} d \theta$$

which equals $$\frac{1}{4} \int_{0}^{2 \pi}\frac{\cos 2 \theta}{1+2\cos^{2} \theta} d \theta$$

I cannot figure out how to easily switch the bounds of the given integral to the interval $[0,2 \pi]$ is there any method to easily do this?Can help explain how to do this easily and for any given interval?

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For $$\int_0^{\pi} \frac{8}{5+2\cos \theta} d\theta$$ note that that the only variable function is $\cos \theta$. Draw the line $x=\pi$ on the graph of $\cos x$, to notice that it has reflectional symmetry about it. In other words, $\cos(\pi + x) = \cos(\pi-x)$. This means the area under the graph in $[0,\pi]$ equals that in $[\pi, 2\pi]$ and the result follows.

Notice that $\frac{1}{1+ \sin^2x}$ is even and so $$\int_{-\pi}^{\pi} \frac{1}{1+\sin^2x}=2 \int_0^{\pi} \frac{1}{1+\sin^2x}$$ Again, by the same reflectional symmetry at $x=\pi$ if we expand the interval to $[0,2\pi]$, we would have to divide by $2$ to preserve the area.

For the last one, you can use a trig identity and the integral becomes$$\int_0^{\frac{\pi}{2}} \frac{\cos 2\theta}{2+\cos 2\theta} d\theta$$

This time, $\cos 2\theta$ has reflectional symmetry along $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$, leading to the areas under the graph in $[0,\frac{\pi}{2}],\, [\frac{\pi}{2}, \pi],\, [\pi, \frac{3\pi}{2}],\, [\frac{3\pi}{2}, 2\pi]$ all being equal.

Note: In general, you might find these properties useful.

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  • $\begingroup$ So when considering these areas under the graphs, we do not consider areas below the x-axis as being negative? $\endgroup$ – user707991 Apr 22 '20 at 20:54
  • $\begingroup$ We do, obviously. $\endgroup$ – Tavish Apr 22 '20 at 21:04

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