Calculus: Analytic Geometry and Calculus, with Vectors

(^520) Exponential and logarithmic functions
situations in which matters are simplified by inserting a cl which is
different from 0.
Letting
(9.55) I1 = f xe y dx,
we set
u=x, dv=eydx
du=1dx, v = fetdx= e-
and conclude that

I1 = -xe z + f e y dx = -xe z - e y + c.

The same idea is useful when n is a positive integer and

(9.56)

Setting

I = f xne z dx.

u=xn, dv=eydx

du = nxn-1 dx, v = fey dx = -e-y

gives the reduction formula

I _ - xne y + n f xn-1e y dx

which expresses In in terms of I._,. In particular,

f x2e-y dx = -x2e y - 2xe y - 2e- + c.

If n is a positive integer and

(9.57)

setting

J. = fxn sin x dx,

u = xn, de = sinxdx

du = nxn-1 dx, v = f sin x dx = - cos x

gives the formula

J. = -xn cos x + n f xn-1 cos x dx.

If n = 1, the last integral is easily evaluated. In case n > 1, we can

integrate by parts again. Setting

it = xn-1, de = cos x dx

du = (n - 1)xn-2 dx, v = f cos x dx = sin x

gives

J. = -xn cos x + nxn-1 sin x - n(n - 1) f xn-2 sin x dx.

In particular,

fx2sinxdx= -x2cosx+2xsinx+2cosx+c.