Integration By Parts -- From Wolfram MathWorld

Home » How To Integrate By Parts » Integration By Parts -- From Wolfram MathWorld

Maybe your like

Integration by parts is a technique for performing indefinite integration intudv or definite integration int_a^budv by expanding the differential of a product of functions d(uv) and expressing the original integral in terms of a known integral intvdu. A single integration by parts starts with

d(uv)=udv+vdu, (1)

and integrates both sides,

intd(uv)=uv=intudv+intvdu. (2)

Rearranging gives

intudv=uv-intvdu. (3)

For example, consider the integral intxcosxdx and let

u=x dv=cosxdx (4)
du=dx v=sinx, (5)

so integration by parts gives

intxcosxdx=xsinx-intsinxdx (6)
=xsinx+cosx+C, (7)

where C is a constant of integration.

The procedure does not always succeed, since some choices of u may lead to more complicated integrals than the original. For example, consider again the integral intxcosxdx and let

u=cosx dv=xdx du=-sinxdx v=1/2x^2, (8)

giving

intxcosxdx=1/2x^2cosx-1/2intx^2(-sinx)dx (9)
=1/2x^2cosx+1/2intx^2sinxdx, (10)

which is more difficult than the original (Apostol 1967, pp. 218-219).

Integration by parts may also fail because it leads back to the original integral. For example, consider intx^(-1)dx and let

u=x dv=x^(-2)dx du=dx v=-x^(-1), (11)

then

intx^(-1)dx=-1-int(-x^(-1))dx+C=intx^(-1)dx+C-1, (12)

which is same integral as the original (Apostol 1967, p. 219).

The analogous procedure works for definite integration by parts, so

int_a^budv=[uv]_a^b-int_a^bvdu, (13)

where [f]_a^b=f(b)-f(a).

Integration by parts can also be applied n times to intf^((n))(x)g(x)dx:

u=g(x) dv=f^((n))(x)dx du=g^'(x)dx v=f^((n-1))(x). (14)

Therefore,

intf^((n))(x)g(x)dx=g(x)f^((n-1))(x)-intf^((n-1))(x)g^'(x)dx. (15)

But

intf^((n-1))(x)g^'(x)dx=g^'(x)f^((n-2))(x)-intf^((n-2))(x)g^('')(x)dx (16)
intf^((n-2))(x)g^('')(x)dx=g^('')(x)f^((n-3))(x)-intf^((n-3))(x)g^((3))(x)dx, (17)

so

intf^((n))(x)g(x)dx=g(x)f^((n-1))(x)-g^'(x)f^((n-2))(x) +g^('')(x)f^((n-3))(x)-...+(-1)^nintf(x)g^((n))(x)dx. (18)

Now consider this in the slightly different form intf(x)g(x)dx. Integrate by parts a first time

u=f(x) dv=g(x)dx du=f^'(x)dx v=intg(x)dx, (19)

so

intf(x)g(x)dx=f(x)intg(x)dx-int[intg(x)dx]f^'(x)dx. (20)

Now integrate by parts a second time,

u=f^'(x) dv=intg(x)dx du=f^('')(x)dx v=intintg(x)(dx)^2, (21)

so

intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+int[intintg(x)(dx)^2]f^('')(x)dx. (22)

Repeating a third time,

intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+f^('')(x)intintintg(x)(dx)^3-int[intintintg(x)(dx)^3]f^(''')(x)dx. (23)

Therefore, after n applications,

intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+f^('')(x)intintintg(x)(dx)^3-...+(-1)^(n+1)f^((n))(x)int...int_()_(n+1)g(x)(dx)^(n+1)+(-1)^nint[int...int_()_(n+1)g(x)(dx)^(n+1)]f^((n+1))(x)dx. (24)

If f^((n+1))(x)=0 (e.g., for an nth degree polynomial), the last term is 0, so the sum terminates after n terms and

intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+f^('')(x)intintintg(x)(dx)^3-...+(-1)^(n+1)f^((n))(x)int...int_()_(n+1)g(x)(dx)^(n+1). (25)

Tag » How To Integrate By Parts