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Question
Evaluate : `int_0^1 x tan^-1x*dx`
Solution
Let I = `int_0^1 x tan^-1x*dx`
= `int_0^1 (tan^-1 x)(x)*dx`
= `[(tan^-1x) int x*dx]_0^1 - int_0^1[d/dx(tan^-1x)* int x*dx]*dx`
= `[(x^2tan^-1x)/2]_0^1 -int_0^1 (1)/(1 + x^2)*x^2/(2)*dx`
= `((1^2tan^-1 1)/2 - 0) - (1)/(2) int_0^1 (1 + x^2 - 1)/(1 + x^2)*dx`
= `(pi/4)/(2) - (1)/(2) int_0^1 (1 - 1/(1 + x^2))*dx`
= `pi/(8) - (1)/(2)[x - tan^-1(x)]_0^1`
= `pi/(8) - (1)/(2)[(1 - tan^-1 1) - 0]`
= `pi/(8) - (1)/(2)(1 - pi/4)`
= `pi/(8) - (1)/(2) + pi/(8)`
= `pi/(4) - (1)/(2)`.
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