$\int_{1 / 2}^{1} \frac{d x}{x \sqrt{4 x^{2}-1}}$

Solve:

$\begin{array}{l} \int_{1 / 2}^{1} \frac{d x}{x \sqrt{4 x^{2}-1}} \\ =\int_{1 / 2}^{1} \frac{2 d x}{2 x \sqrt{(2 x)^{2}-1}}=\left[\sec ^{-1}(2 x)\right]_{1 / 2}^{1} \\ =\sec ^{-1} 2-\sec ^{-1} 1=\frac{\pi}{3}-0=\frac{\pi}{3} \end{array}$

এবং $x=2 \cos \theta$. তাহলে $d x=-2 \sin \theta d \theta$

Limit: $x=1$হলে $\theta=\cos ^{-1} \frac{1}{2}=\frac{\pi}{3}$ এবং

$\begin{aligned} & x=2 \text { if } \theta=\cos ^{-1} 1=0 \\ \therefore \quad & I=\int_{\pi / 3}^{0} \frac{-2 \sin \theta d \theta}{4 \cos ^{2} \theta \sqrt{4\left(1-\cos ^{2} \theta\right)}} \\ & =\int_{\pi / 3}^{0} \frac{-2 \sin \theta d \theta}{4 \cos ^{2} \theta \cdot 2 \sin \theta}=-\frac{1}{4} \int_{\pi / 3}^{0} \sec ^{2} \theta d \theta \end{aligned}$

$\begin{array}{l} =-\frac{1}{4}[\tan \theta]_{\pi / 3}^{0}=-\frac{1}{4}\left(\tan 0-\tan \frac{\pi}{3}\right) \\ =-\frac{1}{4}(0-\sqrt{3})=\frac{\sqrt{3}}{4} \end{array}$