evaluate the integral, give an exact answer $int_{0}^{1}\frac{dx}{sqrt{25 - 4x^{2}}}$

Explanation:

Step1: Use substitution

Let $x = \frac{5}{2}\sin\theta$, then $dx=\frac{5}{2}\cos\theta d\theta$. When $x = 0$, $\theta=0$; when $x = 1$, $\sin\theta=\frac{2}{5}$, so $\theta=\arcsin(\frac{2}{5})$. The integral $\int_{0}^{1}\frac{dx}{\sqrt{25 - 4x^{2}}}$ becomes $\int_{0}^{\arcsin(\frac{2}{5})}\frac{\frac{5}{2}\cos\theta d\theta}{\sqrt{25-4\times(\frac{5}{2}\sin\theta)^{2}}}$.

Step2: Simplify the denominator

Simplify $\sqrt{25-4\times(\frac{5}{2}\sin\theta)^{2}}=\sqrt{25 - 25\sin^{2}\theta}=\sqrt{25(1 - \sin^{2}\theta)} = 5\cos\theta$ (since $1-\sin^{2}\theta=\cos^{2}\theta$ and $\cos\theta\geq0$ for the range of $\theta$ we are considering). The integral is now $\int_{0}^{\arcsin(\frac{2}{5})}\frac{\frac{5}{2}\cos\theta d\theta}{5\cos\theta}=\frac{1}{2}\int_{0}^{\arcsin(\frac{2}{5})}d\theta$.

Step3: Evaluate the integral

$\frac{1}{2}\int_{0}^{\arcsin(\frac{2}{5})}d\theta=\frac{1}{2}[\theta]_{0}^{\arcsin(\frac{2}{5})}=\frac{1}{2}\arcsin(\frac{2}{5})$.

Answer:

$\frac{1}{2}\arcsin(\frac{2}{5})$