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प्रश्न
Evaluate: `int_0^(1/2) 1/((1 - 2x^2) sqrt(1 - x^2)) "d"x`
उत्तर
Let I = `int_0^(1/2) 1/((1 - 2x^2) sqrt(1 - x^2)) "d"x`
Put x = sin θ
∴ dx = cos θ dθ
When x = 0, θ = sin−10 = 0 and
When x = `1/2`, θ = `sin^-1 (1/2) = pi/6`
∴ I = `int_0^(pi/6) 1/((1 - 2sin^2theta)(sqrt(1 - sin^2theta))) cos theta "d"theta`
= `int_0^(pi/6) 1/((cos 2theta)(cos theta)) cos theta "d"theta`
= `int_0^(pi/6) 1/(cos 2theta) * "d"theta`
= `int_0^(pi/6) sec 2theta "d"theta`
= `1/2 [log|sec 2theta + tan 2theta|]_0^(pi/6)`
= `1/2[log|sec2(pi/6) + tan2(pi/6)|] - 1/2[log|sec2(0) + tan2(0)|]`
= `1/2[log|sec(pi/3) + tan(pi/3)|] - 1/2[log|sec(0) + tan(0)|]`
= `1/2[log|2 + sqrt(3)|] - 1/2 log|1|`
∴ I = `1/2 log|2 + sqrt(3)|`
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