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प्रश्न
Evaluate the following : `int x^2*cos^-1 x*dx`
उत्तर
Let I = `int x^2.cos^-1 x*dx`
= `int (cos^-1x)*x^2dx`
= `(cos^-1x) int x^2*dx- int d/dx(cos^-1x) int x^2*dx]*dx`
= `(cos^-1x) (x^3/3) - int ((-1)/sqrt(1 - x^2)) (x^3/3)*dx`
= `x^3/(3) cos^-1x + (1)/(3) int (x^2.x)/sqrt(1 - x^2)*dx`
In `int x^3/sqrt(1 - x^2)*dx`, put 1 – x2 = t
∴ – 2x.dx= dt
∴ x.dx = `-(1)/(2)dt`
Also, x2 = 1 – t
∴ I = `x^3/(3) cos^-1x + (1)/(3) int ((1 - t))/sqrt(t) (-1/2)*dt`
= `x^3/(3) cos^-1x - (1)/(6) int (1/sqrt(t) - sqrt(t))*dt`
= `x^3/(3) cos^-1x - (1)/(6) int t^(-1/2) dt + (1)/(6) int t^(1/2)*dt`
= `x^3/(3) cos^-1x - (1)/(6) (t^(1/2)/(1/2)) + (1)/(6) t^(3/2)/(3/2) + c`
= `x^3/(3) cos^-1x - (1)/(3)sqrt(1 - x^2) + (1)/(9)(1 - x^2)^(3/2) + c`.
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