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प्रश्न
Evaluate the following : `int_0^pi x*sinx*cos^4x*dx`
उत्तर
Let I = `int_0^pi x*sinx*cos^4x*dx` ...(1)
We use the property, `int_0^a f(x)*dx = int_0^a f(a - x)*dx`
Here a = π.
Hence changing x by π – x, we get
I = `int_0^pi (pi - x)*sin(pi - x)*[cos(pi - x)]^4*dx`
= `int_0^pi (pi - x)*sinx*cos^4x*dx` ...(2)
Adding(1) and (2), we get
2I = `int_0^pi x*sinx*cos^4x*dx + int_0^pi (pi - x)*sinx*cos^4x*dx`
= `int_0^pi (x + pi - x)*sinx*cos^4x*dx`
= `pi int_0^pi sinx*cos^4x*dx`
∴ I = `pi/(2) int_0^pi cos^4x*sinx*dx`
Put cos = t
∴ – sinx · dx = dt
∴ sinx · dx = – dt
When x 0, t = cos 0 = 1
When x = π cos π = – 1
∴ I = `pi/(2) int_1^(-1) t^4(- dt)`
= `- pi/(2) int_(1)^(-1) t^4*dt`
= `- pi/(2)[(t^5)/5]_1^(-1)`
= `- pi/(10)[t^5]_1^(-1)`
= `- pi/(10)[(- 1)^5 - (1)^5]`
= `- pi/(10) (- 1 - 1)`
= `(2pi)/(10)`
= `pi/(5)`.
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