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प्रश्न
Evaluate the following:
`int_0^(1/2) ("d"x)/((1 + x^2)sqrt(1 - x^2))` (Hint: Let x = sin θ)
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उत्तर
Let I = `int_0^(1/2) ("d"x)/((1 + x^2)sqrt(1 - x^2))`
Put x = sin θ
∴ dx = cos θ dθ
Changing the limits, we get
When x = 0
∴ sin θ = θ
∴ θ = 0
When x = `1/2`
∴ sin θ = `1/2`
∴ θ = `pi/6`
∴ I = `int_0^(pi/6) (cos theta "d"theta)/((1 + sin^2theta)sqrt(1 - sin^2theta))`
= `int_0^(pi/6) (cos theta "d"theta)/((1 + sin^2theta) costheta)`
= `int_0^(pi/6) 1/(1 + sin^2theta) "d"theta`
Now, dividing the numerator and denominator by cos2θ, we get
= `int_0^(pi/6) (1/cos^2theta)/(1/(cos^2theta) + (sin^2theta)/(cos^2theta)) "d"theta`
= `int_0^(pi/6) (sec^2theta)/(sec^2theta + tan^2theta) "d"theta`
= `int_0^(pi/6) (sec^2theta)/(1 + tan^2theta + tan^2theta) "d"theta`
= `int_0^(pi/6) (sec^2theta)/(2tan^2theta + 1) "d"theta`
Put tan θ = t
∴ sec2θ dθ = t
Changing the limits, we get
When θ = 0
∴ t = tan 0 = 0
When θ = `pi/6`
∴ t = `tan pi/6 = 1/sqrt(3)`
∴ I = `int_0^(1/sqrt(3)) "dt"/(2"t"^2 + 1)`
= `1/2 int_0^(1/sqrt(3)) "dt"/("t"^2 + 1/2)`
= `1/2 int_0^(1/sqrt(3)) "dt"/("t"^2 + (1/sqrt(2))^2)`
= `1/2 xx 1/(1/sqrt(12)) [tan^-1 "t"/(1/sqrt(12))]_0^(1/sqrt(3))`
= `1/sqrt(2) tan^-1 [sqrt(2)"t"]_0^(1/sqrt(3)`
= `1/sqrt(2) [tan^-1 sqrt(2)/sqrt(3) - tan^-1 0]`
= `1/sqrt(2) tan^-1 sqrt(2/3)`
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