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Question
Evaluate the following : `int_0^1 t^5 sqrt(1 - t^2)*dt`
Solution
Let I = `int_0^1 t^5 sqrt(1 - t^2)*dt`
Put t = sin θ
∴ dt = cos θ dθ
When t = 1, θ = sin–11 = `pi/(2)`
When t = 0, θ = sin–10 = 0
∴ I = `int_0^(pi/2) sin^5 theta sqrt(1 - sin^2 theta)cos theta*d theta`
I = `int_0^(pi/2) sin^5 theta*cos theta* cos theta*d theta`
= `int_0^(pi/2) sin^5 theta(1 - sin^2 theta)*d theta`
= `int_0^(pi/2) (sin^5 theta - sin^7 theta)*d theta`
= `int_0^(pi/2) sin^5 theta*d theta - int_0^(pi/2) sin^7 thetad theta`.
Using Reduction formula, we get
I = `4/5*2/3 - 6/7*4/5*2/3`
= `(8)/(15)[1 - 6/7]`
= `(8)/(15) xx (1)/(7)`
= `(8)/(105)`.
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