Definite integration

其實我也很好奇答案是什麼OA O
所以我去找了，所以答案是找來的(爆)

這個積分如果是不定積分，那會長得很難看
不過如果像這樣是特定範圍積分，以下有一個特異的解法：

∫_(0 to π) xsinx/(2 + cosx) dx
= ∫_(x = 0 to π) -x/(2 + cosx) dcosx (變數代換)
= ∫_(x = 0 to π) -x d[ln(cosx+2)] (變數代換)
= -x * [ln(cosx+2)] (x = 0 to π) + ∫_(x = 0 to π) ln(cosx+2) dx (分部積分)
= 0 + ∫_(0 to π) ln(cosx+2) dx

接下來的解法就神奇了...

設I(k) = ∫_(0 to π) ln(1 + kcosx) dx (加入了一個參數，k必須在0到1之間ln才不會爆)
dI(k)/dk = ∫_(x = 0 to π) d[ln(1 + kcosx)]/dk dx
= ∫_(x = 0 to π) cosx/(1 + kcosx) dx
= ∫_(x = 0 to π) (1/k + cosx)/(1 + kcosx) - (1/k)/(1 + kcosx) dx
= ∫_(x = 0 to π) 1/k dx - (1/k)∫_(x = 0 to π) 1/(1 + kcosx) dx

然後設t = tan(x/2)，所以x = 2arctant

= π/k - (1/k) ∫_(t = 0 to infinite) 1/(1 + k[(1-t^2)/(1+t^2)]) d(2arctant)
= π/k - (1/k) ∫_(t = 0 to infinite) [2/(1+t^2)] / [1 + k(1-t^2)/(1+t^2)] dt
= π/k - (1/k) ∫_(t = 0 to infinite) 2/[(1+k)+(1-k)t^2] dt
= π/k - (1/k)[2/(1+k)] * [√(1+k)/(1-k)] arctan{[√(1+k)/(1-k)]t} (t = 0 to infinite)
= π/k - (1/k)[2/(1+k)] * [√(1+k)/(1-k)] [π/2]
= π/k - π/[k√(1-k^2)]

所以dI(k)/dk = π/k - π/[k√(1-k^2)]
得到I(k) = I(0) + ∫_(k = 0 to k) π/k - π/[k√(1-k^2)] dk
= πln[1/2 + [√(1-k^2)]/2]

所以原題目∫_(0 to π) xsinx/(2 + cosx) dx
= ∫_(0 to π) ln(cosx+2) dx
= ∫_(0 to π) ln2 + ln(1 + (1/2)cosx) dx
= πln2 + I(1/2) = πln(1+√3/2)


累死了，不過這個方法真的很神奇XD\
我也不知道有沒有打錯，請參考下面的網頁，往下拉有解答圖片OA O