Symmetry of trig integrals = $0$? $\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}$

calculus

$$\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}dx$$

Because the integral limits go from $-\pi$ to $\pi$ …can one immediately conclude this integral equals zero? The area under the curve should cancel itself out, across the $y$-axis, right?

Best Answer

You can certainly use symmetry, since if $$ f(x) = \frac{\cos(x)}{\sqrt{4+3\sin(x)}},$$ then $$ f(x-\pi/2) = \frac{\sin (x)}{\sqrt{4-3 \cos (x)}},$$ which is an odd function. Thus, $$ \int_{-\pi}^{\pi} f(x-\pi/2) = 0. $$ Geometrically, this means that $f$ has odd symmetry about the line $x=\pi/2$, thus $$ \int_{-\pi/2}^{3\pi/2} f(x) dx = 0, $$ as you can see:

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Of course, we then get that the integral is zero over any interval of length $2\pi$ by periodicity.

You can also use periodicity directly since, with substitution $u=\sin(x)$, we get $$ \int_{-\pi}^{\pi} \frac{\cos(x)}{\sqrt{4+3\sin(x)}} \, dx = \int_0^0 \frac{1}{\sqrt{4+3u}} \, du = 0. $$ In fact, this shows that $$ \int_I g(\sin(x))\cos(x) \, dx = 0, $$ for any integrable function $g$ over any interval $I$ of length $2\pi$.

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