Question

Evaluate: `int_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2)  "d"x`

Answer 1

Let I = `int_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2)  "d"x`

Put sin⁻¹x = t

∴ x = sin t

∴ dx = cos t dt

When x = 0, t = 0 and when x = `1/sqrt(2)`, t = `pi/4`

∴ I = `int_0^(pi/4) "t"/(1 - sin^2"t")^(3/2) xx cos "t"  "dt"`

= `int_0^(pi/4) "t"/(cot^2 "t")^(3/2) xx cos "t"  "dt"`

= `int_0^(pi/4) "t" sec^2 "t"  "dt"`

= `["t" int sec^2 "t"  "dt"]_0^(pi/4) - int_0^(pi/4)["d"/("dt") ("t") int sec^2 "t"  "dt"]"dt"`

= `["t"*tan "t"]_0^(pi/4) - int_0^(pi/4)1* tan "t"  "dt"`

= `(pi/4* tan  pi/4 - 0) - [log|sec "t"|]_0^(pi/4)`

= `pi/4(1) - [log|sec  pi/4| -log|sec 0|]`

= `p/4 - (log sqrt(2) - log 1)`

= `pi/4 - (log 2^(1/2) - 0)`

∴ I = `pi/4 - 1/2 log 2`