I am trying to compute the integral <mtable displaystyle="true"> <mlabeledtr> <mtd id=

Abram Boyd

Abram Boyd

Answered question

2022-06-13

I am trying to compute the integral
(1) 0 π / 3 x cos ( x ) d x

Answer & Explanation

Nia Molina

Nia Molina

Beginner2022-06-14Added 21 answers

I = 0 π 3 x cos ( x ) d x
Making the substitution x = π 2 t
I = π 2 π 6 π 2 d t sin t π 6 π 2 t d t sin t = I 1 I 2
Next, we make the substitution d t sin t = d ( ln ( tan t 2 ) )
I 1 = π 2 π 6 π 2 d ( ln ( tan t 2 ) ) = π 2 ln ( tan π 12 )
Using tan x 2 = 1 cos x sin x and ln ( 2 3 ) = ln ( 2 + 3 )
I 1 = π 2 ln ( 2 + 3 )
I 2 = π 6 π 2 t d t sin t = t ln ( tan t 2 ) | π 6 π 2 π 6 π 2 ln ( tan t 2 ) d t
= π 6 ln ( 2 + 3 ) 0 π 2 ln ( tan t 2 ) d t + 0 π 6 ln ( tan t 2 ) d t
In the second term we make the substitution tan x = t, and the third term was evaluated
G = β ( 2 ) - Catalan's constant)
I 2 = π 6 ln ( 2 + 3 ) 2 0 1 ln t 1 + t 2 d t 4 3 G = π 6 ln ( 2 + 3 ) + 2 G 4 3 G
I 2 = π 6 ln ( 2 + 3 ) + 2 3 G
I = I 1 I 2 = π 3 ln ( 2 + 3 ) 2 3 G

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