積分の公式 (3回目)

2025/10/25(土)
積分の公式 (3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
y = tanh(x) = {exp(x)-exp(-x)}/{exp(x)+exp(-x)}
exp(2x)-1 = yexp(2x)+y
exp(2x)(1-y) = y+1
exp(2x) = (1+y)/(1-y)
2x = log|(1+y)/(1-y)|
tanh^-1(y) = (1/2)log|(1+y)/(1-y)|
 
 
■ 微分
▼ 逆双曲線関数の微分
2{tanh^-1(x)}' = {log|(1+x)/(1-x)|}'
= {(1+x)/(1-x)}'{(1-x)/(1+x)}
= [(1+x)'{1/(1-x)}+{(1+x){1/(1-x)}']{(1-x)/(1+x)}
= [{1/(1-x)}+{(1+x)(-1){-1/(1-x)²}]{(1-x)/(1+x)}
= [{1/(1-x)}+{(1+x)/(1-x)²}]{(1-x)/(1+x)}
= {1+(1+x)/(1-x)}/(1+x)
= {(1-x)+(1+x)}/{(1-x)(1+x)}
= 2/(1-x²)
{tanh^-1(x)}' = 1/(1-x²)
 
▼ 逆双曲線関数の微分の応用
{tanh^-1(sinx)}' = (sinx)'{1/(1-sin²x)} = cosx(1/cos²x)
= 1/cosx
 
{tanh^-1(cosx)}' = (cosx)'{1/(1-cos²x)}
= -sinx(1/sin²x)
= -1/sinx
 
{tanh^-1(tanx)}' = (tanx)'{1/(1-tan²x)}
= (sinx/cosx)'/(1-tan²x)
= {cosx/cosx+sinx(1/cosx)'}/(1-tan²x)
= (cos²x/cos²x+sin²x/cos²x)/(1-tan²x)
= (1/cos²x)/(1-tan²x) = 1/(cos²x-sin²x)
= 1/cos2x
 
 
■ 積分
▼ 逆双曲線関数
tanh^-1(x) = (1/2)log|(1+x)/(1-x)|
 
▼ 積分公式1
∫{1/(1-x²)}dx = tanh^-1(x) + C = (1/2)log|(1+x)/(1-x)| + C
を検証
1/(1-x) + 1/(1+x) = {(1+x)+(1-x)}/{(1+x)(1-x)} = 2/(1-x²)
∫{1/(1-x²)}dx = (1/2){∫{1/(1-x)}dx + ∫{1/(1+x)}dx}
= (1/2)(-log|1-x| + log|1+x|) + C
= (1/2)(log|(1+x)/(1-x)| + C
 
▼ 積分公式2
∫(1/cosx)dx = tanh^-1(sinx) + C = (1/2)log|(1+sinx)/(1-sinx)| + C
を検証
1/cosx = cosx/cos²x = cosx/(1-sin²x)
t = sinxと置くと、dt/dx = cosx、dx = dt/cosx
∫(1/cosx)dx = ∫{cosx/(1-sin²x)}dx
= ∫{1/(1-t²)}dt  … 積分公式1より
= (1/2)log|(1+t)/(1-t)| + C = (1/2)log|(1+sinx)/(1-sinx)| + C
 
▼ 積分公式3
∫(1/sinx)dx = tanh^-1(-cosx) + C = (1/2)log|(1-cosx)/(1+cosx)| + C
= log|tan(x/2)| + C
を検証
1/sinx = sinx/sin²x = sinx/(1-cos²x)
t = cosxと置くと、dt/dx = -sinx、dx = dt/(-sinx)
∫(1/sinx)dx = -∫{sinx/(1-cos²x)}dx
= ∫{1/(1-t²)}dt  … 積分公式1より
= (1/2)log|(1+t)/(1-t)| + C = (1/2)log|(1+cosx)/(1-cosx)| + C
 
▼ 積分公式4
∫(1/cos2x)dx = tanh^-1(tanx) + C = (1/2)log|(1+tanx)/(1-tanx)| + C
= (1/4)log|(1+sin2x)/(1-sin2x)| + C
を検証
t = 2xと置くと、dt/dx = 2、dx = dt/2
∫(1/cos2x)dx = (1/2)∫(1/cost)dt  … 積分公式2より
= (1/2)(1/2)log|(1+sint)/(1-sint)| + C
= (1/4)log|(1+sin2x)/(1-sin2x)| + C
 
 
■ 結果
▼ 定義域
xの定義域に注意
 
▼ 逆双曲線関数
tanh^-1(x) = (1/2)log|(1+x)/(1-x)|
 
▼ 微分公式
{tanh^-1(x)   }' =  2/(1-x²)
{tanh^-1(sinx)}' =  1/cosx
{tanh^-1(cosx)}' = -1/sinx
{tanh^-1(tanx)}' =  1/cos2x
 
▼ 積分公式
∫{1/(1-x²)}dx = tanh^-1(x)    +C = (1/2)log|(1+x)/(1-x)|        +C
∫(1/cosx)dx   = tanh^-1( sinx)+C = (1/2)log|(1+sinx)/(1-sinx)|  +C
∫(1/sinx)dx   = tanh^-1(-cosx)+C = log|tan(x/2)|                +C
∫(1/cos2x)dx  = tanh^-1( tanx)+C = (1/4)log|(1+sin2x)/(1-sin2x)|+C