14. 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 $\theta=\arcsin\frac{2}{5}$.

Step2: Rewrite the integral

Substitute $x$ and $dx$ into the integral: [ \begin{align*} \int_{0}^{1}\frac{dx}{\sqrt{25 - 4x^{2}}}&=\int_{0}^{\arcsin\frac{2}{5}}\frac{\frac{5}{2}\cos\theta d\theta}{\sqrt{25-4\times(\frac{5}{2}\sin\theta)^{2}}}\ &=\int_{0}^{\arcsin\frac{2}{5}}\frac{\frac{5}{2}\cos\theta d\theta}{\sqrt{25 - 25\sin^{2}\theta}}\ &=\int_{0}^{\arcsin\frac{2}{5}}\frac{\frac{5}{2}\cos\theta d\theta}{5\sqrt{1-\sin^{2}\theta}} \end{align*} ] Since $1-\sin^{2}\theta=\cos^{2}\theta$, the integral becomes $\int_{0}^{\arcsin\frac{2}{5}}\frac{\frac{5}{2}\cos\theta d\theta}{5\cos\theta}=\frac{1}{2}\int_{0}^{\arcsin\frac{2}{5}}d\theta$.

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:

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