Question

Evaluate the following : `int_0^1 (cos^-1 x^2)*dx`

Answer 1

Let I = `int_0^1 (cos^-1 x^2)*dx`

Put cos^–1x = t
∴ x = cos t
∴ dx = – sin t ·dt

When x = 0, t = cos^–10 = `pi/(2)`

When x = 1, t = cos^–11 = 0

∴ I = `int_(pi/2)^0 t^2*(- sin t)*dt`

= ` -int_(pi/2)^0 t^2sin t *dt`

= `int_0^(pi/2) t^2 sint*dt         ...[because int_a^b f(x)*dx = -int_b^a f(x)*dx]`

= `[t^2 int sint*dt]_0^(pi/2) - int_0^(pi/2)[d/dx(t^2) int sint*dt]*dt`

= `[t^2 ( cos t)]_0^(pi/2) - int_0^(pi/2) 2t*(- cos t)*dt`

= `[- t^2cos t]_0^(pi/2) + 2int_0^(pi/2) t*cos t*dt`

= `[ - pi/4 cos  pi/2 + 0] + 2{[t int cos t*dt]_0^(pi/2) - int_0^(pi/2)[d/dt (t) int cos t*dt]*dt}`

= `0 + 2{[t sin t]_0^(pi/2) - int_0^(pi/2) 1*sin t*dt}  ...[because cos  pi/2 = 0]`

= `2[t sin t]_0^(pi/2) - 2[(- cos t)]_0^(pi/2)`

= `2[pi/2 sin  pi/2 - 0] - 2[- cos   pi/2 + cos 0]`

= `2[pi/2 xx 1] - 2[- 0 + 1]`

= π – 2.