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