Question

`int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx` = ______ 

Options

• `pi/4 + 1/2log2`

• `pi/4 - 1/2log2`

• `pi/2 + log2`

• `pi/2 - log2`

Answer 1

`int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx` = `underline(pi/4 - 1/2log2)`.

Explanation:

Let I = `int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx`

Put `sin^-1x =  t ⇒ 1/sqrt(1 - x^2)dx = dt`

∴ `I = int_0^{pi/4} t sec^2t dt`

= `[t . tant]_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|]`

= `pi/4 - (logsqrt2 - log 1)`

= `pi/4 - (log2^{1/2} - 0)`

∴ I = `pi/4 - 1/2log2`