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Sum

Evaluate the following : `int_0^1 (1/(1 + x^2))sin^-1((2x)/(1 + x^2))*dx`

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#### Solution

Let I = `int_0^1 (1/(1 + x^2))sin^-1((2x)/(1 + x^2))*dx`

Put x = tan t, i.e. t = tan^{–1}x

∴ dx = sec^{2}t dt

When x = 1, t = tan^{–1}1 = `pi/(4)`

When x = 0, t = tan–1 0 = 0

∴ I = `int_0^(pi/4) (1/(1 + tan^2t))sin^-1 ((2tan t)/(1 + tan^2t))sec^2t*dt`

= `int_0^(pi/4) (1)/(sec^2t) sin^-1 (sin 2t) sec^2t*dt`

= `int_0^(pi/4) 2t*dt`

= `2int_0^(pi/4)t*dt`

= `2[(t^2)/2]_0^(pi/4)`

= `2[pi/(32) - 0]`

= `pi^2/(16)`.

Concept: Fundamental Theorem of Integral Calculus

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