Discussion Forum

If $\int_{0}^{k} \frac{dx}{2+18x^{2}}=\frac{\pi}{24}$  , then the value of k is 

Answer:

Explanation:

We have

$\int_{0}^{k} \frac{dx}{2+18x^{2}}=\frac{\pi}{24}$

= $\frac{1}{18}\int_{0}^{k} \frac{dx}{(\frac{1}{3})^{2}+x^{2}}=\frac{\pi}{24}$

 $\Rightarrow$   $\frac{1}{18}\times\frac{1}{\frac{1}{3}}[tan^{-1}3x]_0^k=\frac{\pi}{24}$

$\Rightarrow$   $[tan^{-1}3k-tan^{-1}0]=\frac{\pi}{4}$

$\Rightarrow$   $tan^{-1}3k=\frac{\pi}{4}$

$\Rightarrow$   $3k=tan\frac{\pi}{4}$

$\Rightarrow$     3k=1

$\Rightarrow$   k=$\frac{1}{3}$