I am trying to evaluate the integration for $x$, where $a$ and $b$ are just constants:

$$\int ^a _0 \ \sqrt{a^2 - x^2} \cos (bx) dx $$

but nothing seems to be working. I have attempted integrating it by parts and substitution, but unless I have done some mistake, it led to nothing.

When trying it by parts it results in something like:

$$(a^2 - x^2)^{1/2} \frac{1}{b} \sin (bx) - \int \frac{1}{b} \sin(bx) (-x) (a^2 - x^2) ^{-1/2}$$

which in itself is sort of a loop. Any help would be greatly appreciated.

PS: An approximation would do.

  • $\begingroup$ What is $I$ here? $\endgroup$ Mar 29, 2023 at 9:15
  • $\begingroup$ I is unecessary, it was just how I labelled the integrals., tbh it is better if I remove it $\endgroup$ Mar 29, 2023 at 9:17
  • $\begingroup$ But then you no longer have an equation, just an integral, so there is nothing to solve for. The integral must be set equal to something to solve for $a$, otherwise it's not clear what you're asking, or maybe I am missing something here. Or maybe you mean "evaluate the integral in terms of $a$" instead of "solve the equation for $a$"? $\endgroup$ Mar 29, 2023 at 9:22
  • $\begingroup$ You can consider $I$ is a measured value. $\endgroup$ Mar 29, 2023 at 9:23
  • 1
    $\begingroup$ What is $b$? Just a constant? $\endgroup$ Mar 29, 2023 at 9:25

1 Answer 1


Divide through: $$F(a,b)=a^2\int_0^a\sqrt{1-x^2/a^2}~\cos(ab~x/a)~\frac{\mathrm dx}{a}$$ Substitute $t=x/a$: $$F(a,b)=a^2\int_0^1 \sqrt{1-t^2}~\cos(abt)\mathrm dt$$ So now consider $ab=z$ and consider

$$f(z)=\int_0^1 \sqrt{1-t^2}~\cos(zt)\mathrm dt$$ Your desired integral $F(a,b)$ can be given as $a^2 f(ab)$.

Note that $f$ is even, e.g $f(-z)=f(z)$. So WLOG consider $z\geq 0$. (Note also that $f(0)=\pi/4$.)

This is a known integral, identity 10.9.4 reads $$\frac{2}{\sqrt \pi}~\frac{(z/2)^\nu}{\Gamma(\nu+1/2)}\int_0^1(1-t^2)^{\nu-1/2}\cos(zt)\mathrm dt = J_{\nu}(z) \\ \Re \nu>\frac{-1}{2}$$

Taking $\nu=1$, we get

$$\frac{2}{\sqrt{\pi}}~\frac{z/2}{\Gamma(3/2)}~\int_0^1 \sqrt{1-t^2}\cos(zt)=J_1(z) \\ f(z)=\int_0^1\sqrt{1-t^2}\cos(zt)\mathrm dt=\frac{\sqrt{\pi}~\Gamma(3/2)}{z}~J_1(z) \\ =\frac{\pi}{2}\frac{J_1(z)}{z}$$

As a sanity check, because $J_1$ is odd, $$f(-z)=\frac{\pi}{2}~\frac{J_1(-z)}{-z}=\frac{\pi}{2}~\frac{-J_1(z)}{-z}=\frac{\pi}{2}~\frac{J_1(z)}{z}=f(z)$$

And also, $$f(0)=\lim_{z\to 0 }\frac{\pi}{2}\frac{J_1(z)}{z}=\frac{\pi}{2}\frac{1}{2}=\frac{\pi}{4}$$


For small $z$, use the Taylor series $$J_1(z)=\frac{1}{2}z-\frac{1}{16}z^3+\frac{1}{384}z^5+\mathrm O(z^7) \\ \text{as}~z\to 0$$

And for large $z$, use the asymptotic expansion $$J_1(z)\asymp \sqrt{\frac{2}{\pi z}}\cos\left(z-\frac{3\pi}{4}\right)+\sqrt{\frac{2}{\pi z}}\mathrm e^{|\Im z|}\mathrm O(|z|^{-1}) \\ \text{as}~z\to\infty$$

Altogether, $$F(a,b)=\int ^a _0 \ \sqrt{a^2 - x^2} \cos (bx) \mathrm dx \\ =\begin{cases}\frac{\pi}{2}\frac{a}{b}J_1(ab) & b\neq 0 \\ a^2\frac{\pi}{4} & b=0\end{cases}$$

  • $\begingroup$ This is way above my knowledge of maths, so I have some questions. How is the gamma function, $\Gamma$, put here? And why was $\mu =1$, why not some other value? $\endgroup$ Mar 29, 2023 at 10:17
  • $\begingroup$ If you put $\nu =1$ in the identity I gave, $(1-t^2)^{\nu-1/2}$ becomes $\sqrt{1-t^2}$ which is what we want in the integral. $\endgroup$
    – K.defaoite
    Mar 29, 2023 at 14:20
  • $\begingroup$ The Gamma function just appears in the referenced integral identity. DLMF should give a reference that you can look at. $\endgroup$
    – K.defaoite
    Mar 29, 2023 at 14:21
  • $\begingroup$ I also made use of the fact that $\Gamma(3/2)=\Gamma(1+1/2)=\frac{1}{2}\Gamma(1/2)=\sqrt \pi /2.$ $\endgroup$
    – K.defaoite
    Mar 29, 2023 at 14:22
  • $\begingroup$ Yes, this integral is found in section 9.1.20 of Abramowitz & Stegun. $\endgroup$
    – K.defaoite
    Mar 30, 2023 at 10:29

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