Page 117 - Schaum's Outline of Theory and Problems of Advanced Calculus

108 INTEGRALS [CHAP. 5

1 u 2
2
2
Then cos x ¼ cos x=2 sin x=2 ¼ :
1 þ u 2
1 2 du
2
2
Also du ¼ sec x=2 dx or dx ¼ 2cos x=2 du ¼ :
2 1 þ u 2
ð
du 1 1 1 1
Thus the integral becomes ¼ tan 1 u=2 þ c ¼ tan tan x=2 þ c:
2
u þ 4 2 2 2
ð x sin x
5.22. Evaluate dx.
2
0 1 þ cos x
Let x ¼ y. Then
x sin x ð yÞ sin y sin y y sin y
ð ð ð ð
I ¼ 2 dx ¼ 2 dy ¼ 2 dy 2 dy
0 1 þ cos x 0 1 þ cos y 0 1 þ cos y 0 1 þ cos y
ð
2

1
¼ dðcos yÞ I ¼ tan ðcos yÞj 0 I ¼ =2 I
2
0 1 þ cos y
2
2
i.e.; I ¼ =2 I or I ¼ =4:
=2 sin x
ð p ffiffiffiffiffiffiffiffiffiffi
5.23. Prove that ffiffiffiffiffiffiffiffiffiffiffi dx ¼ .
ffiffiffiffiffiffiffiffiffiffi
p
p
0 sin x þ cos x 4
Letting x ¼ =2 y,we have
ð =2 p ffiffiffiffiffiffiffiffiffiffi ð =2 p ffiffiffiffiffiffiffiffiffiffi ð =2 p ffiffiffiffiffiffiffiffiffiffiffi
sin x cos y cos x
ffiffiffiffiffiffiffiffiffiffi dx
I ¼ p ffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffi dx ¼ p ffiffiffiffiffiffiffiffiffi dy ¼ p
p p ffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi
0 sin x þ cos x 0 cos y þ sin y 0 cos x þ sin x
Then
ð =2 p ffiffiffiffiffiffiffiffiffiffi ð =2 p cos x
ffiffiffiffiffiffiffiffiffiffiffi
sin x
ffiffiffiffiffiffiffiffiffiffi dx
p ffiffiffiffiffiffiffiffiffiffi p
I þ I ¼ p ffiffiffiffiffiffiffiffiffiffiffi dx þ p ffiffiffiffiffiffiffiffiffiffiffi
0 sin x þ cos x 0 cos x þ sin x
=2 cos x =2
ð p ffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi ð
sin x þ
¼ p ffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffi dx ¼ dx ¼
p
0 sin x þ cos x 0 2
from which 2I ¼ =2 and I ¼ =4.
The same method can be used to prove that for all real values of m,
ð =2 m
sin x
m
m
sin x þ cos x dx ¼ 4
0
(see Problem 5.89).
Note: This problem and Problem 5.22 show that some definite integrals can be evaluated without first
finding the corresponding indefinite integrals.
NUMERICAL METHODS FOR EVALUATING DEFINITE INTEGRALS
1 dx
ð
5.24. Evaluate 2 approximately, using (a) the trapezoidal rule, (b) Simpson’s rule, where the
0 1 þ x
interval ½0; 1Š is divided into n ¼ 4 equal parts.
2
Let f ðxÞ¼ 1=ð1 þ x Þ. Using the notation on Page 98, we find x ¼ðb aÞ=n ¼ð1 0Þ=4 ¼ 0:25.
Then keeping 4 decimal places, we have: y 0 ¼ f ð0Þ¼ 1:0000, y 1 ¼ f ð0:25Þ¼ 0:9412, y 2 ¼ f ð0:50Þ¼ 0:8000,
y 3 ¼ f ð0:75Þ¼ 0:6400, y 4 ¼ f ð1Þ¼ 0:50000.
(a) The trapezoidal rule gives