Question

If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is

• A. 3
• B. 4
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{1}{3}$

Answer 2

We have, $\int\limits_{0}^{k} \frac{d x}{2+18 x^{2}}=\frac{\pi}{24}$
$\Rightarrow \frac{1}{18} \int\limits_{0}^{k} \frac{d x}{\left(\frac{1}{3}\right)^{2}+x^{2}}=\frac{\pi}{24}$
$\Rightarrow \frac{1}{18} \times \frac{1}{\frac{1}{3}}\left[\tan ^{-1} 3 x\right]_{0}^{k}=\frac{\pi}{24}$
$\Rightarrow \left[\tan ^{-1} 3 k-\tan ^{-1} 0\right]=\frac{\pi}{4}$
$\Rightarrow \tan ^{-1} 3 k=\frac{\pi}{4}$
$\Rightarrow 3 k=\tan \frac{\pi}{4}$
$\Rightarrow 3 k=1$
$\Rightarrow k=\frac{1}{3}$