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For solving the integral: $$ \int_a^b \sqrt{\alpha^2 - \beta^2 x^2} \, dx $$ I've been taught to use $x = \frac{\alpha}{\beta} \sin(t)$ in order to get $$ \frac{\alpha^2}{\beta} \int_{\arcsin(a \beta/\alpha)}^{\arcsin(b \beta/\alpha)} \sqrt{1-\sin(t)^2} \cos(t) \,dt $$ which is easier since it is $\int\cos^2$ and by the identity $\cos(t)^2 = \frac{1}{2}(1 + \cos(2t))$ it's done. But what if $x \geq \alpha/\beta$? Am I missing something? Thanks and sorry if I am asking something obvious.

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If $x > \alpha/\beta$ (assuming that $\alpha,\beta > 0$), then the integrand $\sqrt{\alpha^2 - \beta^2x^2}$ will be complex, since $\alpha^2 - \beta^2 x^2 < \alpha^2 - \beta^2\cdot\frac{\alpha^2}{\beta^2} = 0$. The function can still be integrated, but I assume your domain of integration will avoid such problems, because you would need knowledge of integration of complex functions to deal with the integral if that ever happens.

If $x = \alpha/\beta$ on the nose for some $x$ in the domain of integration, then you're still fine using ordinary methods to evaluate the integral, because $\sqrt{\alpha^2 - \beta^2x^2} = \sqrt{\alpha^2 - \beta^2\cdot\frac{\alpha^2}{\beta^2}} = \sqrt{0} = 0$.

Hence, if you want your integral to end up being a real number, your $a$ and $b$ will have to satisfy $$ a,b\in\left[-\left|\frac{\alpha}{\beta}\right|,\left|\frac{\alpha}{\beta}\right|\right]. $$

You can also see that you can only consider the substitution above when $a$ and $b$ are as above by looking at what happens to the limits under the substitution: $\arcsin x$ is only defined on $\left[-1,1\right]$, so we need $a\beta/\alpha \geq -1$ and $b\beta/\alpha\leq 1$ (here I'm again assuming you're taking $\alpha,\beta > 0$) in order for the second equation to even make sense! But you can see that the conditions just stated are the same as $a$ and $b$ being in the interval $\left[-\left|\frac{\alpha}{\beta}\right|,\left|\frac{\alpha}{\beta}\right|\right]$. So, we see that the substitution $x = \frac{\alpha}{\beta}\sin t$ is only valid when $-\left|\frac{\alpha}{\beta}\right|\leq a\leq x\leq b\leq \left|\frac{\alpha}{\beta}\right|$.

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