∫ 0 π / 2 ln ( 4 + 3 cos x 4 + 3 sin x ) d x = ?
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5 seconds! @brian charlesworth ∫ a b f ( x ) = ∫ a b f ( a + b − x ) This would be a better approach I guess.
Haha. Well, yours and Azhaghu's brains work three times faster than mine, apparently. :) I suppose you're right, but I just wanted to provide a more expansive explanation.
C'mon! You can't compare me with yourself. I am just a beginner. You are the real math champ! ⌣ ¨
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The integral can be rewritten as
∫ 0 2 π lo g ( 4 + 3 sin ( x ) ) d x − ∫ 0 2 π lo g ( 4 + 3 cos ( x ) ) d x = L − M .
For the first of these integrals, namely L , make the substitution y = 2 π − x . Then d y = − d x and sin ( x ) = sin ( 2 π − y ) = cos ( y ) . We then have that
L = ∫ 0 2 π lo g ( 4 + 3 sin ( x ) ) d x = − ∫ 2 π 0 lo g ( 4 + 3 cos ( y ) ) d y =
∫ 0 2 π lo g ( 4 + 3 cos ( y ) ) d y .
Now since y is just a dummy variable at this point we can call it anything we want. So letting it be x , we see that in fact L = M and so the desired solution is 0 .