Question

If $\underset{0}{\overset{K}{\int }}\frac{dx}{2+18x^2}=\frac{\pi }{24}$, then the value of K is

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

Answer 1

Answer: (C)

We have, $\underset{0}{\overset{k}{\int }}\frac{dx}{2+18x^2}=\frac{\pi }{24}$
$\Rightarrow \frac{1}{18}\underset{0}{\overset{k}{\int }}\frac{dx}{{\left(\frac{1}{3}\right)}^2+x^2}=\frac{\pi }{24}$
$\Rightarrow \frac{1}{18}\times \frac{1}{\frac{1}{3}}{\left[{\tan }^{-1}3x\right]}_0^k=\frac{\pi }{24}$
$\Rightarrow \left[{\tan }^{-1}3k-{\tan }^{-1}0\right]=\frac{\pi }{4}$
$\Rightarrow {\tan }^{-1}3k=\frac{\pi }{4}$
$\Rightarrow 3k=\tan \frac{\pi }{4}$
$\Rightarrow 3k=1$
$\Rightarrow k=\frac{1}{3}$