Post a New Question

Pure Mathematics

posted by on .

In=∫ x^nSinx dx....with the limits π/2 and 0
i found that In would be equal to n(π/2)^(n-1) -n(n-1) In-2
how do i deduce the value of I4? where n=4

  • Pure Mathematics - ,

    you need to do integration by parts.

    ∫x^4 sinx dx
    u=x^4
    du = 4x^3 dx

    dv = sinx dx
    v = -cosx

    ∫x^4 sinx = -x^4 cosx + 4∫x^3 cosx dx

    Now do all that again for each resulting integral, and the powers of x descends until you are left with just ∫sinx dx

    Then just add all the pieces together.

    Hmmm. I think you can save yourself some work. At x=0 all the terms with powers of x and sinx vanish.

    at x=π/2 all the terms with cosx vanish (odd powers of x).

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question