Integration by Parts



 
òf(x)g¢(x)dx = f(x)g(x)-òf¢(x)g(x)dx

In order to find òx5cos x dx, the integration by parts formula follows the steps

òx5cos x dx = x5sin x-ò5x4sin x dx

-ò5x4sin x dx = 5x4cos x-ò20x3cos x dx

-ò20x3cos x dx = -20x3sin x+ò60x2sin x dx

ò60x2sin x dx = -60x2cos x+ò120x cos x dx

ò120x cos x dx = 120x sin x-ò 120sin x dx

-ò 120sin x dx = 120cos x

to obtain

òx5cos x dx = x5sin x+5x4cos x-20x3sin x-60x2cos x+120x sin x+120cos x+C.

A moment's reflection shows that the result was obtained by taking successive derivatives of x5 and taking successive integrals of cos x and combine the result as indicated below:

x5
 
cos x 
5x4
 
sin x 
20x3
 
-cos x 
60x2
 
-sin x 
120x 
 
cos x 
120 
 
sin  x 
 
-cos  x
The procedure can be justified from

ò fg  dx  = fg1-òf¢g1dx = fg1-f¢g2+òf¢¢g2dx = ···

= fg1-f¢g2+f¢¢g3-···+(-1)n-1f(n-1)gn+(-1)nòf(n)gndx

where

gk = ògk-1dx.

  òp(x)exp(x)dx

In view of the above formula we see that if p is a polynomial of degree n then

òp(x)exp(x)dx = [p(x)-p¢(x)+p¢¢(x)-p(3)(x)+···+(-1)np(n)(x)]exp(x).

  Table of Integrals for òxnexp(x)dx

The integration by parts formula yields

òxnexp(x)dx = xnexp(x)-ònxn-1exp(x)dx.

Thus if

òxnexp(x)dx = pn(x)exp(x),

then the polynomial pn(x) is obtained from the previous polynomial pn-1(x) by multiplying by n and adding the term xn.

Exercise: Build up the following table of coefficients for the pn¢s.

Exercise: Compute òln(x)ndx for n = 1,2,···,10.

  òp(x)cos x, òp(x)sin x dx

It is clear now that if p(x) is a polynomial of degree n then

òp(x)cos x = [p(x)-p¢¢(x)+p(4)(x)-p(6)(x)+···]cos x

+[p¢(x)-p(3)(x)+p(5)(x)-p(7)(x)+···]sin x

Table of Integrals for

òxncos x dx, òxnsin x dx

a. Compute òxncos x dx for n = 1,2,3,4,5.

b. Write down the reduction formula for òxncos x dx and òxnsin x dx.

c. Compute the polynomials appeared in òxncos x dx and in òxnsin x dx for n = 1,2,···,14.

  Chebyshev Polynomials

a. Establish the identity
cos(n+1)q = 2cos q cos nq-cos(n-1)q

b. Set cos q = x, cos kq = Tk(x). Show that

T0(x) º 1, T1(x) = x and  Tn+1(x) = 2xTn(x)-Tn-1(x) for  n ³ 1. 

c. List the coefficients of Tn for n = 0,1,2,...,12.

d. Draw the graphs of T1, T2, T3,...,T6 on the square [-1,1]×[-1,1].

e. Calculate the coefficients ck in the expansion 2nxn = åk = 0nckTk(x) for n = 1,2,¼,12.

f. Find the coefficients ak in the integral

òcosnq dq = a0q+ n
å
k = 1
aksin kq
for n = 1,2,···,12.

  òsinnq dq

a. Establish the identities
2sinx cos y = sin(y+x)-sin(y-x) 
2sinx sin y = cos(y-x)-cos(y+x) 

b. Deduce from the identities

2sin2q = 1-cos 2q
4sin3q = 3sin q-sin 3q
8sin4q = 3-4cos 2q+cos 4q
16sin5q = 10sin q-5sin 3q+sin 5q
32sin6q = 10-15cos 2q+6cos 4q-cos 6q

c. Compute the coefficients ak and bk in the expansions

22nsin2n+1q n
å
k = 0
aksin kq
22n-1sin2nq n
å
k = 0
bkcos kq
for n = 1,2,···,10.

d. Compute òsinnq dq for n = 1,2,···,20.

  òtannq dq

The reduction formula follows from

òtannq dq = òtann-2q(sec2q-1)dq = òtann-2sec2qdq-òtann-2q dq

= [(tann-1q)/(n-1)]-òtann-2q dq.

Hence

tan2nq dq = åk = 1n(-1)k-1[(tan2n-2k+1q)/(2n-2k+1)]+(-1)nq

òtan2n+1q dq = åk = 1n(-1)k-1[(tan2n-2k+1q)/(2n-2k+1)]+(-1)nln(cos q).

  òsecnq dq

For even n, write òsec2kqdq = ò(1+tan2q)k-1sec2qdq

to reduce it to ò(1+u2)k-1du and expand the binomial expression. For odd n, set u = secn-2q, dv = sec2qdq and when tan2q appears, use the identity tan2q = sec2q-1.

Example. Find òsec3qdq.

Write u = sec q, dv = sec2qdq

du = sec q tan q dq, v = tan q

òsec3qdq = òsec2q sec qdq = òudv = uv - òvdu = sec q tan q- òtan2q sec q dq

= sec q tan q- ò(sec2q-1) sec q dq = sec q tan q- òsec3q dq+òsec q dq.

Therefore

2òsec3q = sec q tan q+ln|sec q+tan q|

and so

òsec3q = [sec q tan q+ln|sec q+tan q|]/2.