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watch5826c

watch5826c

Answered question

2022-06-16

Evaluating 0 π ln ( 1 + cos x ) d x

Answer & Explanation

feaguelaBapzo

feaguelaBapzo

Beginner2022-06-17Added 9 answers

0 π log ( 1 + cos x ) d x = 0 π / 2 log ( 1 + cos x ) d x + 0 π / 2 log ( 1 + cos ( π x ) ) d x (1) = 0 π / 2 log ( sin 2 x ) d x = 0 π log ( sin x ) d x
And by a notorious identity:
(2) k = 1 n 1 sin k π n = 2 n 2 n ,
hence the RHS of (1) can be computed as a Riemann sum:
(3) 0 π log ( sin x ) d x = lim n + π n k = 1 n 1 log sin π k n = π log 2 .
There is also a well-known proof through symmetry:
I = 0 π log ( sin x ) = 2 0 π / 2 log ( sin ( 2 t ) ) d t = 2 0 π / 2 log ( 2 sin t cos t ) d t = π log 2 + 2 0 π / 2 log ( sin t ) d t + 2 0 π / 2 log ( cos t ) d t (4) = π log 2 + 2 I
from which I = π log 2 immediately follows.
minwaardekn

minwaardekn

Beginner2022-06-18Added 6 answers

Notice,
0 π log ( 1 + cos x )   d x = 2 0 π / 2 log ( 1 + cos 2 x )   d x = 2 0 π / 2 log ( 2 cos 2 x )   d x
= 4 0 π / 2 log ( cos x )   d x + 2 log 2 0 π / 2   d x
(1) = 4 I + π log 2
Where,
(2) I = 0 π / 2 log ( cos x )   d x
(3) I = 0 π / 2 log ( sin ( π 2 x ) )   d x = 0 π / 2 log ( sin x )   d x
Adding (2) & (3), one should get
2 I = 0 π / 2 log ( sin x cos x )   d x = 0 π / 2 log ( sin 2 x 2 )   d x = 0 π / 2 log ( sin 2 x )   d x π 2 log 2
2 I = 1 2 0 π log ( sin x )   d x π 2 log 2 = 2 2 0 π / 2 log ( sin x )   d x π 2 log 2 = I π 2 log 2
I = π 2 log 2
, now setting the value of I in (1), one should get
0 π log ( 1 + cos x )   d x = 4 ( π 2 log 2 ) + π log 2 = π log 2

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