Question

यदि `int_0^"a" 1/(1 + 4x^2)"d"x = pi/8` है, तो a = ______

Answer 1

यदि `int_0^"a" 1/(1 + 4x^2)"d"x = pi/8` है, तो a = `underline(1/2)`.

व्याख्या:

दिया गया है कि `int_0^"a" 1/(1 + 4x^2) "d"x = pi/8`

⇒ `1/4 int_0^"a" 1/((1/4 + x^2)) "d"x = pi/8`

⇒ `int_0^pi 1/([(1/2)^2 + x^2]) "d"x = pi/2`

⇒ `1/(1/2) [tan^-1  x/(1/2)]_0^"a" = pi/2`

⇒ `2[tan^-1 2"a" - tan^-1 0] = pi/2`

⇒ `tan^-1 2"a" = pi/4`

⇒ 2a = `tan  pi/4`

⇒ 2a = 1

⇒ a = `1/2`