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Question
Show that `int_0^pi log(1+acos x)/cos x dx=pi sin^-1 a 0 ≤ a ≤1.`
Solution
Let I (a) be the given integral. By the rule of differentiation under the integral sign.
`(DI)/(Da)= int_0^pi (df)/(da) dx=int_0^pi 1/cos x.cos x/(1+acos x)dx=int_0^pi dx/(1+acos x)`
Put `t= tan x/2 , dx=(2dt)/(1+t^2 )and cos x=(1-t^2)/(1+t^2)`
When `x=0, t=0`
When `x= pi, t= tan pi/2= infty`
∴` (dI)/(da) = int_0^infty 1/(1+a((1-t^2)/(1+t^2))).(2 dt)/(1+t^2)`
`(dI)/(da)=int_0^infty (2dt)/((1+t^2)+a(1-t^2))`
`(dI)/(da)=int_0^infty (2dt)/((1+a)+(1-a)t^2)`
`(dI)/(da)=1/(1-a) int_0^infty (2dt)/([(1+a)/(1-a)]+t^2)`
`(dI)/(da)=2/(1-a) sqrt(1-a)/(1+a) [tan^-1 sqrt((1-a)/(1+a)) ]_0^infty`
`(dI)/(da) = 2/sqrt(1-a^2).pi/2`
`(dI)/(da)=pi/sqrt(1-a^2)`
Integrating both sides w.r.t. a, we get
`I=pi sin^-1 a+c`
To find c, put a = 0
`I(0)=pi sin^-1 0+c, c=0`
∴ `I =pi sin^-1 a`
∴` int_0^pi log(1+acos x)/cos x dx=pi sin^-1 a`
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