Question

Evaluate:

`intcos^-1(sqrt(x))dx`

Answer 1

Let I = `intcos^-1(sqrt(x))dx`

Put `sqrt(x)` = t

∴ x = t²

∴ dx = 2t dt

∴ I = `int(cos^-1t)(2)dt`

= `(cos^-1t) int 2tdt - int[d/dt (cos^-1t) int 2tdt]dt`

= `(cos^-1t).(2t^2)/2 - int(-1)/sqrt(1 - t^2).(2t^2)/2dt`

= `t^2.cos^-1t - int(-t^2)/sqrt(1 - t^2)dt`

= `t^2.cos^-1t - int((1 - t^2) - 1)/sqrt(1 - t^2)dt`

= `t^2.cos^-1t - int[sqrt(1 - t^2) - 1/sqrt(1 - t^2)]dt`

= `t^2.cos^-1t - intsqrt(1 - t^2)dt + int1/sqrt(1 - t^2)dt`

= `t^2.cos^-1t - [t/2 sqrt(1 - t^2) + 1/2 sin^-1t] + sin^-1t + c`     ...`[∵ int sqrt(a^2 - x^2)dx = x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1(x/a) + c]`

= `t^2.cos^-1t - t/2 sqrt(1 - t^2) - 1/2 sin^-1t + sin^-1t + c`

= `t^2.cos^-1t - t/2 sqrt(1 - t^2) + 1/2sin^-1t + c`

= `x.cos^-1(sqrt(x)) - sqrt(x)/2 sqrt(1 - x) + 1/2 sin^-1(sqrt(x)) + c`.