$$\int\frac{\mathrm dx}{x^2 \sqrt{16-x^2}}$$
when substituting $\,x=4\sin\theta ,\;\mathrm dx=4\cos\theta\, \mathrm d\theta,\,$ it becomes
$$\int\frac{4\cos\theta\, \mathrm d\theta}{4^2\sin^2\theta\sqrt{16-4^2\sin^2\theta}}$$
then
$$\frac{1}{4}\int\frac{\cos\theta \,\mathrm d\theta}{\sin^2\theta\times4\sqrt{1-\sin^2\theta}}$$
The question is why did we take the positive root of 16 and took it out? as $\sqrt{16} \;is \pm 4 $
Why do we take the positive square root only from $\sqrt{a^2-x^2}$ when integrating using trig substitutions
calculusintegration
Best Answer
You wrote that “$\sqrt{16}$ is $\pm4$”. This is wrong. Yes, the number $16$ has two square roots: $4$ and $-4$. But $\sqrt{16}$ stands for the non-negative root of $16$, which is $4$. For the same reason, $\sqrt{16-4^2\sin^2\theta}=4\sqrt{1-\sin^2\theta}$.