The integral is
$$\int\frac{dx}{\sqrt{4-5x^2}}$$
I was trying:
$$=\int \frac{dx}{\sqrt{5(\frac{4}{5}-x^2)}}$$
$$=\frac{1}{\sqrt{5}}\int \frac{dx}{\sqrt{\frac{4}{5}-x^2}}$$
$$=\frac{1}{\sqrt{5}}
\int \frac{dx}
{\sqrt{\left(\sqrt{\frac{4}{5}}\right)^2-x^2}}$$
Let $x = \sqrt{\frac{4}{5}}\sin\theta$ and $dx = \sqrt{\frac{4}{5}}\cos\theta\, d\theta.$
$$=\frac{\sqrt{5}}{5}\int \frac{d\theta}{\cos\theta}$$
That integral does not have the form of the answer which involves $\arcsin$.
Why this trig substitution does not work?
Best Answer
$$\int \frac{dx}{\sqrt{4-5x^2)}}$$ Samsamradas factored $5$. Instead, I will factor $4$. $$\int \frac{dx}{2\sqrt{1-(\tfrac{\sqrt5x}2)^2)}}$$ Now, let $\frac{\sqrt5 x}2=\sin\theta$ so that $\theta=\arcsin(\tfrac{\sqrt5 x}{2})$. Then, $\frac{\sqrt5 dx}2=\cos\theta\,d\theta$ and hence $dx=\frac2{\sqrt5}\cos\theta\,d\theta.$ The integral becomes $$\int\frac{\frac2{\sqrt5}\cos\theta\,d\theta}{2\sqrt{1-\sin^2\theta}}$$ By using the trigonometric identity $\cos^2\theta+\sin^2\theta=1$, we have $\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}=|\cos\theta|.$ We may assume that $-\frac\pi 2\leq\theta\leq\frac\pi 2$ and hence $|\cos\theta|=\cos\theta.$
We end up with $$\frac1{\sqrt5}\int d\theta=\frac1{\sqrt5}\theta+C\\=\frac1{\sqrt5}\arcsin(\tfrac{\sqrt5 x}2)+C$$