Question

Find:

`intsin^(-1)  sqrt((x)/(a + x))  dx`

Answer 1

Let I = `intsin^(-1)  sqrt((x)/(a + x))  dx`

Put x = a tan² θ

`tan^2θ = x/a`

dx = 2a tan θ sec² dθ

I = `intsin^(-1)sqrt((a tan^2θ)/(a + a tan^2θ))  2a tan θ sec^2 dθ`

= `intsin^(-1)sqrt((a tan^2θ)/(a sec^2θ))  2a tan θ sec^2θ  dθ`

= `sin^(-1)sqrt(sin^2θ)  .  2a tan θ sec^2θ  dθ`

= `intsin^(-1)(sin θ) 2a tan θ sec^2θ  dθ`

= `2a intθ tan θ sec^2 θ  dθ`

Put tan θ = t

θ = tan⁻¹t

sec²θ dθ = dt

I = `2ainttan_(I)^(-1)t_(II) dt`

I = `2a[tan^(-1)t int t  dt - int(d/dttan^(-1)tintt  dt)dt]`

⇒ I = `2a[t^2/2 tan^(-1)t - int1/(1 + t^2)  .  t^2/2 dt]`

⇒ I = `a[t^2 tan^(-1)t - int(t^2 + 1 - 1)/(t^2 + 1)  dt]`

⇒ I = `a[t^2 tan^(-1) t - int(t^2 + 1)/(t^2 + 1)  dt + int(1)/(t^2 + 1)  dt]`

⇒ I = a[t²tan⁻¹ t − t + tan⁻¹ t] + C

⇒ I = a[θ tan² θ − tan θ + θ] + C

but `tan^2 θ = x/a`

`θ = tan^(-1)sqrt(x/a)`

⇒ I = `a[x/a tan^(-1) sqrt(x/a) - sqrt(x/a) + tan^(-1) sqrt(x/a)] + C`

I = `x tan^(-1) sqrt(x/a) - sqrt(xa) + a tan^(-1)sqrt(x/a) + C`

I = `(x + a)tan^(-1) sqrt(x/a) - sqrt(xa) + C`