Evaluate the definite integral: `int_(-1/(sqrt(3)))^(1/(sqrt(3)))((x^4)/(1-x^4))cos^(-1)((2x)/(1+x^2))dxdot`

Answer

`(pi)/(12)[pi+3log(2+sqrt(3))-4sqrt(3)]`

Explanation

Let `I=int _(-1//sqrt(3))^(1//sqrt(3))((x^(4))/(1-x^(4)))cos^(-1)((2x)/(1+x^(2)))dx` . . . (i)
Put `x=-y rArr dx =- dy`
` :. I=int _(1//sqrt3)^(-1//sqrt(3))(y^(4))/(1-y^(4))cos ^(-1) ((-2y)/(1+y^(2))(-1)dx`
Now , `cos^(-1)(-x)=pi-cos^(-1)x " for" - 1 lex le1`.
`:. I = int _(-1//sqrt(3))^(1//sqrt(3))(y^(4))/(1-y^(4))[pi-cos^(-1)((2y)/(1+y^(2)))]dy`
` = piint_(-1//sqrt(3))^(1//sqrt(3))(y^(4))/(1-y^(4))dy - int _(-1//sqrt(3))^(1//sqrt(3))(y^(4))/(1-y^(4))cos^(-1)((2y)/(1+y^(2)))dy`
`=piint_(-1//sqrt(3))^(1//sqrt(3))(x^(4))/(1-x^(4))dx- int _(-1//sqrt3)^(1//sqrt(3))(x^(4))/(1-x^(4))cos^(-1)((2x)/(1+x^(2)))dx`
` rArr I= piint_(-1//sqrt(3))^(1//sqrt(3))(x^(4))/(1-x^(4))dx-I` " " [ from Eq . (i)]
`rArr 2I=pi int _(-1//sqrt(3))^(1//sqrt(3))(x^(4))/(1-x^(4))dx = pi int_(-1//sqrt(3))^(1//sqrt(3))[-1+(1)/(1-x^(4))]dx`
`=-int _(-1//sqrt(3))^(1//sqrt(3))1dx+pi int_(-1//sqrt(3))^(1//sqrt(3))(dx)/(1-x^(4))`
` =-pi [x] _(-1//sqrt(3))^(1//sqrt(3))+pi I_(1), " where "I_(1), = int_(-1// sqrt(3))^(1//sqrt(3))(dx)/(1-x^(4))`
`rArr 2I =-pi ((1)/(sqrt(3))+(1)/(sqrt(3)))+pi I_(1)=-(2pi)/(sqrt(3))+piI_(1)`
Now , `I _(1)=int _(-1//sqrt(3))^(1//sqrt(3)) (dx)/(1-x^(4))=2 int _(0)^(1//sqrt(3))(dx)/(1-x^(4))`
[ since , the integral is an even function]
`=int_(0)^(1//sqrt(3))(1+1+x^(2)-x^(2))/((1-x^(2))(1+x^(2)))` dx
`= int _(0)^(1//sqrt(3))(1)/(1-x^(2))dx +int _(0)^(1//sqrt(3))(1)/(1+x^(2))dx`
`=int _(0)^(1//sqrt(3))(1)/((1-x)(1+x))dx +int _(0)^(1//sqrt(3))(1)/((1+x^(2)))dx`
`=(1)/(2)int_(0)^(1//sqrt(3))(1)/(1-x)dx+(1)/(2)int_(0)^(1//sqrt(3))(1)/(1+x)dx+int_(0)^(1//sqrt(3))(1)/(1+x^(2))` dx
`=[ -(1)/(2)"in" | 1-x|+(1)/(2) " in" |1+x|+tan^(-1)x]_(0)^(1//sqrt(3))`
`=(1)/(2)[ " In " |(1+x)/(1-x)|]_(0)^(1//sqrt(3)) +[ tan ^(-1)x] _(0)^(1//sqrt(3))`
`=(1)/(2) " In" |(1+1//sqrt(3))/(1-1//sqrt(3))|+"tan"^(-1)(1)/(sqrt(3))`
`=(1)/(2) " In " |(sqrt(3)+1)/(sqrt(3)-1)|+(pi)/(6)=(1)/(2) " In" |((sqrt(3)+1)^(2))/(3-1)|+(pi)/(6)`
`=(1)/(2) "In" (2+sqrt(3))+(pi)/(6)`
`:. 2I =(-2pi)/(sqrt(3))+(pi)/(2) "In" (2+sqrt(3))+(pi^(2))/(6)`
`=(pi)/(6)[pi+3 " In" (2+sqrt(3))-4sqrt(3)]`
`rArr I=(pi)/(12)[pi+3 " In" (2+sqrt(3)-4sqrt(3)]`
Alternate Solution
Since , `cos ^(1)y=(pi)/(2)sin^(-1)y`
`:. cos^(-1)((2x)/(1+x^(2)))=(pi)/(2)-sin^(-1)(2x)/(1+x^(2))=(pi)/(2)-2 tan ^(-1)x`
`I=int_(-1//sqrt(3))^(1//sqrt(3))[(pi)/(2)*(x^(4))/(1-x^(4))-(x^(4))/(1-x^(4))2 tan ^(-1)x] dx`
`[:' (x^(4))/(1-x^(4))2tan^(-1)` x is an odd function ]
`:. I=2(pi)/(2)int_(0)^(1/sqrt(3))(-1+(1)/(1-x^(4)))dx+0`
`=(pi)/(2)int_(0)^(1/sqrt(3))(-2+(1)/(1-x^(2))+(1)/(1+x^(2)))dx`
`=(pi)/(2)[-2x+(1)/(2*1)"log" (1+x)/(1-x)+tan^(-1)x]_(0)^(1//sqrt(3))`
`=(pi)/(2)[-(2)/(sqrt(3))+(1)/(2) "log" (sqrt(3)+1)/(sqrt(3-1))+(pi)/(6)]`
`=(pi)/(12)[pi=3) log (2+sqrt(3)-4sqrt(3)]`