Question

Evaluate the following : `int_0^(pi/2) 1/(6 - cosx)*dx`

Answer 1

Let I = `int_0^(pi/2) 1/(6 - cosx)*dx`

Put `tan(x/2)` = t

∴ x = 2 tan^–1 t

∴ dx = `(2dt)/(1 + t)`
and
cos x = `(1 - t^2)/(1 + t^2)`

When x =  `pi/(2), t = tan(pi/2)` = 1

When x = 0, t = tan 0 = 0

∴ I = `((2dt)/(1 + t^2))/(6 - cos((1 - t^2)/(1 + t^2))`

= `int_0^1 (2dt)/(6(1 + t^2) + 1(1 - t^2)`

= `2 int_0^1 (1)/(t^2 + 7)*dt`

= `2[1/35 tan^-1  t/5]_0^1`

= `2[1/35 tan^-1  1/(3) - 1/(5) tan^-1 0]`

= `(2)/(35) tan^-1  (1)/(3) - (7)/(5) xx 0`

= `(2)/sqrt(35) tan^-1  sqrt(7/5)`.