Page 114 - Schaum's Outline of Theory and Problems of Advanced Calculus

P. 114

CHAP. 5] INTEGRALS 105

Another method:
By Problem 5.10 and Problem 4.3, Chapter 4, it follows that F ðxÞ exists and so FðxÞ must be con-
0
tinuous.

CHANGE OF VARIABLES AND SPECIAL METHODS OF INTEGRATION

5.13. Prove the result (7), Page 95, for changing the variable of integration.
ð x ð t
f fgðtÞg g ðtÞ dt, where x ¼ gðtÞ.
0
Let FðxÞ¼ f ðxÞ dx and GðtÞ¼
a
Then dF ¼ f ðxÞ dx, dG ¼ f fgðtÞg g ðtÞ dt.
0
Since dx ¼ g ðtÞ dt,it follows that f ðxÞ dx ¼ f fgðtÞg g ðtÞ dt so that dFðxÞ¼ dGðtÞ, from which
0
0
FðxÞ¼ GðtÞþ c.
Now when x ¼ a, t ¼ or FðaÞ¼ Gð Þþ c. But FðaÞ¼ Gð Þ¼ 0, so that c ¼ 0. Hence FðxÞ¼ GðtÞ.
Since x ¼ b when t ¼ ,we have
ð b ð
f fgðtÞg g ðtÞ dt
0
f ðxÞ dx ¼
a
as required.
5.14. Evaluate:
p ﬃﬃ
ð ð 1 dx ð 1= 2 x sin 1 x 2
2
ðx þ 2Þ sinðx þ 4x 6Þ dx ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ dx
ðaÞ ðcÞ p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðeÞ p 4
1 ðx þ 2Þð3 xÞ 0 1 x
ð ð ð
cotðln xÞ x 1 x xdx
dx 2 tanh 2 dx
x x þ x þ 1
ðbÞ ðdÞ ð f Þ p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2
2
1
(a) Method 1: Let x þ 4x 6 ¼ u. Then ð2x þ 4Þ dx ¼ du, ðx þ 2Þ dx ¼ du and the integral becomes
2
1 ð 1 1 2
cosðx þ 4x 6Þþ c
2 sin udu ¼ 2 cos u þ c ¼ 2
Method 2:
ð 1 ð 1
2 2 2 2
2 2
ðx þ 2Þ sinðx þ 4x 6Þ dx ¼ sinðx þ 4x 6Þdðx þ 4x 6Þ¼ cosðx þ 4x 6Þþ c
(b)Let ln x ¼ u. Then ðdxÞ=x ¼ du and the integral becomes
ð
cot udu ¼ ln j sin ujþ c ¼ ln j sinðln xÞj þ c
ð ð ð ð
dx dx dx dx
Method 1:
ðcÞ p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ q ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
6 þ x x 2 2 1 2
ðx þ 2Þð3 xÞ 6 ðx xÞ
2
25=4 ðx Þ
1
Letting x ¼ u,this becomes
2
du 1 u 1 2x 1
ð
ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ sin þ c ¼ sin þ c
p 2
25=4 u 5=2 5
Then
ð 1 dx 1 1 1 3

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ sin 1 2x 1 ¼ sin sin 1
p
1 ðx þ 2Þð3 xÞ 5 1 5 5
¼ sin 1 :2 þ sin 1 :6

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