About int_0^(pi/2) arctan(1-sin^2(x)cos^2(x))dx=pi(pi/4-arctan sqrt(sqrt((2)-1)/2)))

Valery Cook

Valery Cook

Answered question

2022-10-29

About 0 π / 2 arctan ( 1 sin 2 ( x ) cos 2 ( x ) ) d x = π ( π 4 arctan 2 1 2 )

Answer & Explanation

Sauppypefpg

Sauppypefpg

Beginner2022-10-30Added 23 answers

0 π / 2 arctan ( 1 cos 2 x sin 2 x ) d x = 0 π / 2 ( π 2 arctan ( 1 1 sin 2 x cos 2 x ) ) d x = π 2 4 0 π / 2 arctan ( sin 2 x ) d x 0 π / 2 arctan ( cos 2 x ) d x = π 2 4 2 0 π / 2 arctan ( cos 2 x ) d x
Consider
I ( a ) = 0 π / 2 arctan ( a cos 2 x ) d x
I ( a ) = 0 π / 2 cos 2 x 1 + a 2 cos 4 x d x = 0 π / 2 sec 2 x sec 4 x + a 2 d x = 0 π / 2 sec 2 x ( 1 + tan 2 x ) 2 + a 2 d x = 0 d t t 4 + 2 t 2 + a 2 + 1 ( tan x = t )
With the substitution t a 2 + 1 t
I ( a ) = 1 a 2 + 1 0 t 2 t 4 + 2 t 2 + a 2 + 1 d t
I ( a ) = 1 2 a 2 + 1 0 a 2 + 1 + t 2 t 4 + 2 t 2 + a 2 + 1 d t = 1 2 a 2 + 1 0 1 + 1 + a 2 t 2 ( t a 2 + 1 t ) 2 + 2 ( 1 + a 2 + 1 ) d t = 1 2 a 2 + 1 d y y 2 + 2 ( 1 + a 2 + 1 ) ( t a 2 + 1 t = y ) = π 2 2 1 1 + a 2 1 + a 2 + 1
Integrating back,
I ( 1 ) I ( 0 ) = I ( 1 ) = π 2 2 0 1 d a 1 + a 2 1 + a 2 + 1 = π 2 2 1 2 d t t 1 ( t + 1 ) ( a 2 + 1 = t ) = π 2 0 2 1 d u u 2 + 2 ( t 1 = u 2 ) = π 2 ( arctan u 2 | 0 2 1 = π 2 arctan ( 2 1 2 )
Hence,
0 π / 2 arctan ( 1 cos 2 x sin 2 x ) d x = π 2 4 π arctan ( 2 1 2 )

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