How to solve the integral $IR_0\int_0^{\sqrt{R^2-R_0^2}}(R_0^2+x^2)^{-\frac{3}{2}}dx$?

Question

This integral is to calculate total illuminance of a circular surface.

$$ I\,R_{0}\int_{0}^{\,\sqrt{\,R^{2} - R_{0}^{2}\,\,}\,}\!\! \left(\vphantom{\Large A}\,R_{0}^{2} + x^{2}\,\right)^{-3/2}\,\,\mathrm{d}x $$

Because the ^$-3/2$ function has another function of $x$ in itself, I can't find a way to calculate the integral.

Someone told me to use integration by parts. but I can't put the derivative of the function inside the $\mbox{^$-3/2$}$ function beside it to use that rule, because it will contain $x$ itself and make a product.

Everything except $x$ is constant.

Can anyone help me ?.

Answer 1

Let $x=R_0\tan(\theta)$ so that $dx=R_0 \sec^2(\theta)\,d\theta$. Then, denoting by $\theta_0$, $\theta_0=\arctan\left(\frac{\sqrt{R^2-R_0^2}}{R_0}\right)$, we have

$$\begin{align} IR_0\int_0^{\sqrt{R^2-R^2_0}}\left(R^2_0+x^2\right)^{-3/2}\,dx&=\frac{I}{R_0} \int_{0}^{\theta_0} \cos(\theta)\,d\theta\\\\ &=\frac{I}{R_0}\sin\left(\theta_0\right)\\\\ &=\frac{I}{RR_0}\sqrt{R^2-R_0^2} \end{align}$$

Answer 2

For $x=R_0\tan{t}$, $dx=\frac{1}{\cos^2{t}}$, we have $$I_R=\int_0^{\arctan{\sqrt{\frac{R^2}{R_0^2}-1}}} \frac{\cos{t}}{R_0^2}=\frac{\sin{t}}{R_0^2}|_0^{\arctan{\sqrt{\frac{R^2}{R_0^2}-1}}}=\frac{\sqrt{R^2-R_0^2}}{RR_0^2},$$ where in the last step we used identity $$\sin{t}=\frac{\tan{t}}{\sqrt{\tan^2{t}+1}}.$$