Re: [積分] 定積分

[Frobenius (▽．(▽×▽φ)=0)]

※ 引述《JULIKEBEN (啾西)》之銘言：
:   兀/4
: ∫       ln(1+tanx)dx
:   0
: 想問怎麼下手
: 請高手指教謝謝

1+tanx = (cosx+sinx)/cosx = [sin(π/2-x)+sinx]/cosx

sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB
───────────────
sin(A+B) + sin(A-B) = 2sinAcosB

A + B = π/2-x
A - B = x

2A = π/2 => A = π/4 => B = π/4-x

sin(π/2-x)+sinx = 2sin(π/4)cos(π/4-x) = 2(√2/2)cos(π/4-x) = √2cos(π/4-x)

1+tanx = √2cos(π/4-x)/cosx

π/4                π/4
∫   ㏑(1+tanx)dx = ∫   ㏑( √2 cos(π/4-x) / cosx )dx
 0                   0

  π/4               π/4                       π/4
= ∫   ㏑( √2 )dx + ∫   ㏑( cos(π/4-x) )dx - ∫   ㏑(cosx)dx
   0                  0                          0

( Let X = π/4-x => dX = -dx，x = 0～π/4 => X = π/4～0 )

                     0                π/4
= (π/4)㏑( √2 ) - ∫   ㏑(cosX)dX - ∫   ㏑(cosx)dx
                    π/4               0

               π/4              π/4
= (π/8)㏑2 + ∫    ㏑(cosx)dx - ∫   ㏑(cosx)dx
               0                  0

= (π/8)㏑2

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→ JULIKEBEN:thanks=)
→ andy74139:sin(A+B)-sin(A-B)應該是2cosAsinB吧？
※ 編輯: Frobenius       來自: 140.109.112.96       (01/03 20:28)
→ Frobenius:sin(A+B)+sin(A-B)=2cosAcosB 已修正為2sinAcosB,感謝^^