if `int_0^k (dx)/(2+8x^2)=pi/16` then find the value of `k`

Options

A. `1/2`
B. `1/3`
C. `1/4`
D. `1/5`

Answer

A

Explanation

Given, `int_(0)^(k) 1/(2 + 8 x^(2))* dx = pi/16`
` :." " 1/2 int_(0)^(k) 1/(1+4x^(2)) * dx = pi/16`
`:. " " 1/2 int_(0)^(k) 1/(4(x^(2)+1/4))* dx = pi/16`
` :. " " 1/8 int_(0)^(k) 1/(x^(2) + (1/2)^(2))* dx = pi/16`
` :. " " 1/8*(1)/((1)/(2))[tan^(-1)((x)/((1)/(2)))]_(0)^(k) = pi/16`
` :. " " 1/4[tan^(-1)(2x)]_(0)^(k)=pi/16`
` :. " " [tan^(-1)(2x)]_(0)^(k) = pi/16 xx 4/1`
` :. " " tan^(-1) 2k - tan^(-1) 0 = pi/4`
` :. " " tan^(-1) 2k - 0 = pi/4`
` :. " " tan^(-1) 2k = pi/4`
` :. " " 2k = tan(pi/4) = 1`
` :. " " k = 1/2 `
Hence, the correct answre from the given alternatives is (a).