# Evaluate the following : ∫01logx1-x2⋅dx - Mathematics and Statistics

Sum

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

#### Solution

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).

Is there an error in this question or solution?