Question

Evaluate the following integrals : `int sqrt((9 - x)/x).dx`

Answer 1

 Let I = `int sqrt((9 - x)/x).dx`

= `int sqrt((9 - x)/x.(9 - x)/(9 - x)).dx`

= `int (9 - x)/sqrt(9x - x^2).dx`

Let 9 – x = `"A"[d/dx (9x - x^2)] + "B"`

= A(9 – 2x) + B
∴ 9 – x = (9A + B) – 2Ax
Comparing the coefficient of x and constant on both the sides, we get
– 2A = – 1 and 9A + B = 9

∴ `"A" = (1)/(2) and 9(1/2) + "B"` = 9

∴ B = `(9)/(2)`

∴ 9 – x = `(1)/(2)(9 - 2x) + (9)/(2)`

∴ I = `int (1/2(9 - 2x) + 9/2)/sqrt(9x - x^2).dx`

= `(1)/(2) int (9 - 2x)/sqrt(9x - x^2).dx + (9)/(2) int (1)/sqrt(9x - x^2).dx`

= `(1)/(2)"I"_1 + (9)/(2)"I"_2`

In I₁, put 9x – x² = t
∴ (9 – 2x)dx = dt

∴ I₁ = `int (1)/sqrt(t)dt`

= `intt^(-1/2)dt`

= `t^(1/2)/(1/2) + c_1`

= `2sqrt(9x - x^2) + c_1`

I₂ = `int(1)/sqrt(81/4 - (x^2 - 9x + 81/4)).dx`

= `int (1)/sqrt((9/2)^2 - (x - 9/2)^2).dx`

= `sin^-1((x - 9/2)/(9/2)) + c_2`

== `sin^-1((2x - 9)/9) + c_2`

∴ I = `sqrt(9x - x^2) + (9)/(2) sin^-1((2x - 9)/9) + c`, where c = c₁ + c₂.