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Question
Evaluate the following : `int_0^pi (sin^-1x + cos^-1x)^3 sin^3x*dx`
Solution
Let I = `int_0^pi (sin^-1x + cos^-1x)^3 sin^3x*dx`
We know that, sin–1x + cos–1x = `pi/(2)`
and
sin 3x = 3 sin x – 4 sin3x
∴ 4sin3x = 3 sin x – sin 3x
∴ sin3x = `3/4 sinx - 1/4 sin3x`
∴ I = `int_0^pi (pi/2)^3[3/4 sin x - 1/4 sin 3x]*dx`
= `pi^3/(8) xx 3/4 int_0^pi sin x*dx - pi^2/(8) xx 1/4 int_0^pi sin3x`
= `(3pi^3)/(32) [- cos pi - ( - cos 0)] - pi^3/(32)[- (cos 3pi)/(3) - ((- cos0)/3)]`
= `(3pi^3)/(32)[1 + 1] - pi^3/(32)[1/3 + 1/3]`
= `(6pi^3)/(32) - (2pi^3)/(96)`
= `(18pi^3 - 2pi^3)/(96)`
= `(16pi^3)/(96)`
= `pi^3/(6)`.
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