Page 288 - DMTH401_REAL ANALYSIS

Real Analysis

Notes The foregoing discussion, of course, does not apply when a = ±1 since the conditions for
differentiability are not met.

Example: Here, we consider the integration of
p 1
2
I = 0 ò (acos x bsin x+ 2 ) dx
2
where both a, b > 0, by differentiating under the integral sign.

p 1
2
Let us first find J = ò 0 acos x bsin x+ 2 dx
2
2
Dividing both the numerator and the denominator by cos x yields
2
p 2 sec x
J = ò 0 a btan x+ 2 dx
1 p 1
= 0 ò 2 d(tanx)
b æ a ö 2
2
+ tan x
ç ÷
è b ø
p 2
1 æ - 1 æ b ö ö p
= ç tan ç tanx ÷ ÷ = ×
a, b è è a øø 2 a, b
0
1
p
The limits of integration being independent of a, J = 0 ò 2 dx gives us
2
2
+
acos x bsin x
2
J ¶ p 2 cos xdx
= - 0 ò
¶ a (acos x bsin x+ 2 ) 2
2
p
Whereas J = gives us
2 ab
J ¶ p
= - ×
¶ a 4 a b
3
Equating these two relations then yields
2
p 2 cos x dx p
0 ò 2 2 2 = 3
(acos x bsin x+ ) 4 a b
J ¶
In a similar fashion, pursuing yields
¶ b
2
p sin x dx p
2
0 ò 2 2 2 = 3
(acos x bsin x+ ) 4 ab
Adding the two results then produces
p 1 p æ 1 1 ö
2
I = 0 ò 2 dx = ç + ÷
2
(acos x bsin x+ 2 ) 4 ab è a b ø
Which is the value of the integral I.

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