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प्रश्न
Evaluate : `int_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2)*dx`
उत्तर
Let I = `int_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2)*dx`
= `int_0^(1/sqrt(2)) (sin^-1x)/((1 - x^2)sqrt(1 - x^2))*dx`
Put sin–1 x = t
∴ `(1)/sqrt(1 - x^2)*dx` = dt
Also, x = sin t
When x = `(1)/sqrt(2), t = sin^-1 (1/sqrt(2)) = pi/(4)`
When x = 0, t = sin–10 = 0
∴ I = `int_0^(pi/4) t/(1 - sin^2t)*dt`
= `int_0^(pi/4) t/(cos^2t)*dt`
= `int_0^(pi/4) t sec^2t*dt`
= `[t int sec^2t*dt]_0^(pi/4) - int_0^(pi/4)[d/dt (t) int sec^2t*dt]*dt`
= `[t tant]_0^(p/4) - int_0^(pi/4) 1*tant*dt`
= `[pi/4 tan pi/4 - 0] -[log |sect|]_0^(pi/4)`
= `pi/(4) - [log(sec pi/4) - log (sec 0)]`
= `pi/(4) - [log sqrt(2) - log 1]`
= `pi/(4) - (1)/(2)log2`. ...[∵ log 1 = 0]
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