Question

If `int_0^h1/(2+8x^2)dx=pi/16 `then find the value of h.

Answer 1

`int_0^h "dx"/(2 + "8x"^2) = pi/16`

`1/8 int_0^h "dx"/((1/2)^2 + "x"^2) = pi/16`

`int_0^"h" "dx"/("x"^2 + "h"^2) = pi/2`

`2  "tan"^-1 (2"x")]_0^"h" = pi/2`

`2 "tan"^-1 2 "h" = pi/2`

`"tan"^-1 "2h" = pi/4`

2h = 1

h = `1/2`