積分の公式 (4回目)

2025/10/30(木)
積分の公式 (4回目)
 
(integral)
 
■ 公式
▼ 三角関数
sin2x = 2sinxcosx
cos2x = cos²x-sin²x
= (1-sin²x)-sin²x = 1-2sin²x
= cos²x-(1-cos²x) = 2cos²x-1
1-cosx = 2sin²(x/2)
1+cosx = 2cos²(x/2)
 
▼ 双曲線関数
sinh(x) = {exp(x)-exp(-x)}/2
cosh(x) = {exp(x)+exp(-x)}/2
tanh(x) = {exp(x)-exp(-x)}/{exp(x)+exp(-x)}
 
▼ 逆双曲線関数1
y = sinh(x) = {exp(x)-exp(-x)}/2
2yexp(x) = exp(2x)-1
exp(2x) - 2yexp(x) - 1 = 0
exp(x) = y±√(y²+1) ≧ 0
x = log{y+√(y²+1)}
sinh^-1(y) = log|y+√(y²+1)|
 
▼ 逆双曲線関数2
y = cosh(x) = {exp(x)+exp(-x)}/2
2yexp(x) = exp(2x)+1
exp(2x) - 2yexp(x) + 1 = 0
exp(x) = y±√(y²-1) ≧ 0
x = log{y+√(y²-1)}  … (y ≧ 1)
cosh^-1(y) = log|y+√(y²-1)|
 
 
■ 微分
▼ 逆双曲線関数の微分1
{sinh^-1(x)}' = {log{x+√(x²+1)}}'
= {x+√(x²+1)}'{1/{x+√(x²+1)}}
= {1+(x²+1)'(1/2)/√(x²+1)}/{x+√(x²+1)}
= {1+x/√(x²+1)}/{x+√(x²+1)}
= {√(x²+1)+x}/{x√(x²+1)+(x²+1)}
= {(x²+1)-x²}/[{√(x²+1)-x}{x√(x²+1)+(x²+1)}]
= 1/{x(x²+1)-x²√(x²+1)+(x²+1)√(x²+1)-x(x²+1)}
= 1/{(-x²+(x²+1)√(x²+1)}
= 1/√(x²+1)
{sinh^-1(x)}' = 1/√(x²+1)
 
▼ 逆双曲線関数の微分1
{cosh^-1(x)}' = {log{x+√(x²-1)}}'
= {x+√(x²-1)}'{1/{x+√(x²-1)}}
= {1+(x²-1)'(1/2)/√(x²-1)}/{x+√(x²-1)}
= {1+x/√(x²-1)}/{x+√(x²-1)}
= {√(x²-1)+x}/{x√(x²-1)+(x²-1)}
= {(x²-1)-x²}/[{√(x²-1)-x}{x√(x²-1)+(x²-1)}]
= -1/{x(x²-1)-x²√(x²-1)+(x²-1)√(x²-1)-x(x²-1)}
= -1/{(-x²+(x²-1)√(x²-1)}
= 1/√(x²-1)
{cosh^-1(x)}' = 1/√(x²-1)
 
 
■ 積分
▼ 公式
https://ulprojectmail.blogspot.com/2025/10/integral-2.html
積分の公式 (2回目)
より
 
∫√(r² + x²)dx = (1/2){x√(r² + x²) + r²tanh^-1(x/√(r² + x²))} + C
∫√(r² - x²)dx = (1/2){x√(r² - x²) + r²tan^-1(x/√(r² - x²))} + C
∫{1/√(r² + x²)}dx = tanh^-1{x/√(r² + x²)} + C
∫{1/√(r² - x²)}dx = tan^-1{x/√(r² - x²)} + C
 
▼ 積分公式1
∫{1/√(x²+1)}dx = sinh^-1(x) + C = log|x+√(x²+1)| + C
を検証
r = 1と置いて
∫{1/√(r² + x²)}dx = tanh^-1{x/√(r² + x²)} + C
∫{1/√(1² + x²)}dx = tanh^-1{x/√(1² + x²)} + C
= tanh^-1{x/√(1 + x²)} + C
= (1/2)log|{1+x/√(1 + x²)}/{1-x/√(1 + x²)}| + C
= (1/2)log|{√(1 + x²)+x}/{√(1 + x²)-x}| + C
= (1/2)log|{√(1 + x²)+x}²/{(1 + x²)-x²}| + C
= (1/2)log|{√(1 + x²)+x}²| + C
= log|√(1 + x²)+x| + C
 
▼ 積分公式2
∫{1/√(x²-1)}dx = cosh^-1(x) + C = log|x+√(x²-1)| + C
を検証
r = iと置いて
∫{1/√(r² + x²)}dx = tanh^-1{x/√(r² + x²)} + C
∫{1/√(i² + x²)}dx = tanh^-1{x/√(i² + x²)} + C
= tanh^-1{x/√(x² - 1)} + C
= (1/2)log|{1+x/√(x² - 1)}/{1-x/√(x² - 1)}| + C
= (1/2)log|{√(x² - 1)+x}/{√(x² - 1)-x}| + C
= (1/2)log|{√(x² - 1)+x}²/{(x² - 1)-x²}| + C
= (1/2)log|-{√(x² - 1)+x}²| + C
= log|√(x² - 1)+x| + C
 
 
■ 結果
▼ 定義域
xの定義域に注意
 
▼ 微分
{sinh^-1(x)}' = 1/√(x²+1)
{cosh^-1(x)}' = 1/√(x²-1)
 
▼ 積分
∫{1/√(x²+1)}dx = sinh^-1(x) + C = log|x+√(x²+1)| + C
∫{1/√(x²-1)}dx = cosh^-1(x) + C = log|x+√(x²-1)| + C