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प्रश्न
Evaluate: `int_0^"a" 1/(x + sqrt("a"^2 - x^2)) "d"x`
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उत्तर
Let I = `int_0^"a" 1/(x + sqrt("a"^2 - x^2)) "d"x`
Put x = a sin θ
∴ dx = a cos θ dθ
When x = 0, θ = 0 and when x = a, θ = `pi/2`
∴ I = `int_0^(pi/2) ("a"costheta "d"theta)/("a"sintheta + sqrt("a"^2 - "a"^2 sin^2 theta))`
= `int_0^(pi/2) ("a"costheta"d"theta)/("a"sintheta + "a"sqrt(1 - sin^2 theta))`
`int_0^(pi/2) (cos theta)/(sin theta + sqrt(cos^2theta)) "d"theta`
∴ I = `int_0^(pi/2) (costheta)/(sintheta + cos theta) "d"theta` .......(i)
∴ I = `int_0^(pi/2) (cos(pi/2 - theta))/(sin(pi/2 - theta) + cos(pi/2 - theta))` .......`[∵ int_0^"a" "f"(x)"d"x = int_0^"a" "f"("a" - x)"d"x]`
∴ I = `int_0^(pi/2) (sintheta)/(costheta + sintheta) "d"theta` .......(ii)
Adding (i) and (ii), we get
2I = `int_0^(pi/2) (costheta)/(sintheta + costheta) "d"theta+ int_0^(pi/2) (sin theta)/(cos theta + sin theta) "d"theta`
= `int_0^(pi/2) (cos theta + sin theta)/(sin theta + cos theta) "d"theta`
= `int_0^(pi/2) "d"theta - [theta]_0^(pi/2)`
= `pi/2 - 0`
∴ I = `1/2 xx pi/2`
∴ I = `pi/4`
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