Question

Evaluate the following : `int_0^1 (logx)/sqrt(1 - x^2)*dx`

Answer 1

Let I = `int_0^1 (logx)/sqrt(1 - x^2)*dx`

Put x = sin θ
∴ dx = cos θ dθ
and
`sqrt(1 - x^2) = sqrt(1 - sin^2 theta) = sqrt(cos^2 theta)` = cos θ

When x = 0, sin θ = 0  ∴ θ = 0
When x = 1, sin θ = 1  ∴ θ = `pi/(2)`

∴ I = `int_0^(pi/2) log sin theta *d theta`

Using the property, `int_0^(2a) f(x)*dx = int_0^(a)[f(x) + f(2a - x)]*dx`, we get

I = `int_0^(pi/4) [log sin theta + log sin (pi/2 - theta)]*d theta`

= `int_0^(pi/4) (log sin theta + log cos theta)* d theta`

= `int_0^(pi/4) log sin theta cos theta* d theta`

= `int_0^(pi/4) log((2 sin theta cos theta)/2)*d theta`

= `int_0^(pi/4) (log sin 2 theta - log 2)*d theta`

= `int_0^(pi/4) log sin 2 theta*d theta - int_0^(pi/4) log 2* d theta`

= I₁ – I₂                                                  ...(Say)

I₂ = `int_0^(pi/4) log 2* d theta`

= `log 2 int_0^(pi/4) 1*d theta`

= `log 2 [theta]_0^(pi/4)`

= `(log 2)[pi/4 - 0]`

= `pi/(4) log 2`

I₁ = `int_0^(pi/4) log sin 2 theta * d theta`

Put 2θ = t. 

Then dθ= `dt/(2)`

When θ = 0, t = 0

When θ = `pi/(4), t = 2(pi/4) = pi/(2)`

∴ I₁ = `int_0^(pi/2) log sin t xx dt/(2)`

= `(1)/(2) int_0^(pi/2) log sin theta* d theta`

= `(1)/(2)"I"     ...[ because int_a^b f(x)*dx = int_a^b f(t)*dt]`

∴ I = `(1)/(2) "I" - pi/(4)log 2`

∴ `(1)/(2)"I" = - pi/(4) log 2`

∴ I = `- pi/(2) log 2`

= `pi/(2) log (1/2)`.