Page 290 - DMTH401_REAL ANALYSIS

Real Analysis

Notes p
Now, when a = , we have, from
2
2 ln(1 cos cosx)+ a æ p ö
p
I(a) = 0 ò cosx dx, I ç ÷ = 0
è
2 ø
Hence,
a
I(a) = ò -a da
p
2
1 a
= - a 2
2 p
2
p 2 a 2
= - ,
8 2
which is the value of the integral I(a).

2p cos q
Example: Here, we consider the integral ò e cos(sin )dq .
q
0
We introduce a new variable f, and rewrite the integral as
2p
f
f( ) = 0 ò e f cos q cos( sin )dq
f
q
2p
f
q
Note that for f = 1, f( ) = f(1) = 0 ò e cos q cos(sin )dq
Thus, we proceed
df 2p ¶
f
= ò (e f cos q cos( sin q ) ) dq
df 0 ¶f
2p f cos q
q
f
q
= ò e (cos cos( sin ) sin sin( sin ))dq
f
q
q
-
0
2 1 ¶
p
f
= ò 0 f ¶q (e f cos q sin( sin q ) ) dq
1 2p f cos q
f
= 0 ò d (e sin( sin ) ) q
f
2p
1
f
= (e f cos q sin( sin ) ) q
f
0
= 0.
df
From the equation for f(f) we can see f(0) = 2p. So, integrating both sides of = 0 with respect
df
to f between the limits 0 and 1, yields
f(1) 1
ò df = 0 ò df = 0
f(0)
 f(1) – f(0) = 0

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