next up previous index
Next: 3.7 解答 Up: 演習問題詳解 Previous: 3.5 解答   索引

3.6 解答

3.6

1.

(a) 無理関数 $\sqrt{\frac{ax + b}{cx + d}}$をtとおくことにより，有理関数に直す．

$\displaystyle \int{x\sqrt{1+x}} dx$ $\displaystyle =$ $\displaystyle \int{(t^2 -1)t(2t)} dt \left(\begin{array}{ll} t = \sqrt{1 + x}とおくと t^2 = 1 + x\\ これより x = t^2 - 1, dx = 2t dt \end{array}\right)$  
  $\displaystyle =$ $\displaystyle 2\int(t^4 - t^2) dt = 2(\frac{t^5}{5} - \frac{t^3}{3}) + c$  
  $\displaystyle =$ $\displaystyle 2[\frac{(1 + x)^{5/2}}{5} - \frac{(1+x)^{3/2}}{3}] + c$  

(b) $t = \sqrt{x}$ とおくと $t^2 = x$より $2t dt = dx$. したがって，

$\displaystyle \int{\frac{\sqrt{x}}{\sqrt{x} - 1}} dx$ $\displaystyle =$ $\displaystyle \int{\frac{t}{t-1} 2t} dt$  
  $\displaystyle =$ $\displaystyle 2\int{\frac{t^2}{t-1}} dt$  

ここで 分子の次数 $>$ 分母の次数 より

$\displaystyle \frac{t^2}{t-1} = t + 1 + \frac{1}{t-1}.$

これより
$\displaystyle \int{\frac{\sqrt{x}}{\sqrt{x} - 1}} dx$ $\displaystyle =$ $\displaystyle 2\int{(t + 1 + \frac{1}{t-1})} dt$  
  $\displaystyle =$ $\displaystyle 2[\frac{t^2}{2} + t + \log{\vert t-1\vert}] + c$  
  $\displaystyle =$ $\displaystyle t^2 + 2t + 2\log{\vert t-1\vert} + c$  
  $\displaystyle =$ $\displaystyle x + 2\sqrt{x} + 2\log{\vert\sqrt{x} - 1\vert} + c$  

(c) $t = \sqrt{1 + e^{x}}$とおくと $t^2 = 1 + e^{x}$．これより $2t dt = e^{x}dx$．ここで被積分関数と$dx$を全て$t$の関数と$dt$で表すと

$\displaystyle dx = \frac{2t dt}{e^{x}} = \frac{2t dt}{t^2 - 1}.$

したがって，
$\displaystyle \int{\frac{dx}{\sqrt{1 + e^{x}}}}$ $\displaystyle =$ $\displaystyle \int{\frac{2t}{t(t^2 - 1)}} dt$  
  $\displaystyle =$ $\displaystyle 2\int{\frac{1}{t^2 -1}} dt \left(\begin{array}{l} \int{\frac{1... ...2}} dx \\ = \frac{1}{2a}\log{\vert\frac{x-a}{x+a}\vert} +c \end{array}\right)$  
  $\displaystyle =$ $\displaystyle 2\cdot \frac{1}{2} \log{\vert\frac{t-1}{t+1}\vert} + c$  
  $\displaystyle =$ $\displaystyle \log{\vert\frac{\sqrt{1+e^x} -1}{\sqrt{1+e^{x} + 1}}\vert} +c$  

(d) $t = \sqrt{x - 1}$とおくと$t^2 = x - 1$．これより $2t dt = dx$．ここで被積分関数と$dx$を全て$t$の関数と$dt$で表すと

$\displaystyle x^{2}\sqrt{x-1}dx = (t^{2} + 1)^{2} (t) (2tdt) = 2t^{6} + 4t^{4} + 2t^{2}$

したがって，
$\displaystyle \int{x^{2}\sqrt{x-1}dx}$ $\displaystyle =$ $\displaystyle \int{2t^{6} + 4t^{4} + 2t^{2}} dt$  
  $\displaystyle =$ $\displaystyle \frac{2t^{7}}{7} + \frac{4t^{5}}{5} + \frac{2t^{3}}{3} + c$  
  $\displaystyle =$ $\displaystyle \frac{2(x-1)^{7/2}}{7} + \frac{4(x-1)^{5/2}}{5} + \frac{2(x-1)^{3/2}}{3} + c$  

(e) $t = \sqrt{\frac{x+1}{x-1}}$とおくと $t^2 = \frac{x+1}{x-1}$． ここで$dx$を求めるには，上の式を一旦$x$について解く必要がある．

$\displaystyle t^2$ $\displaystyle =$ $\displaystyle \frac{x+1}{x-1} より$  
$\displaystyle xt^2 - t^2$ $\displaystyle =$ $\displaystyle x + 1$  
$\displaystyle x(t^2 - 1)$ $\displaystyle =$ $\displaystyle t^2 + 1となるので$  
$\displaystyle x$ $\displaystyle =$ $\displaystyle \frac{t^2 + 1}{t^2 - 1}$  

これより
$\displaystyle dx$ $\displaystyle =$ $\displaystyle \frac{2t(t^2 - 1) - (t^2 + 1)(2t)}{(t^2 - 1)^2} dt$  
  $\displaystyle =$ $\displaystyle \frac{-4t}{(t^2 - 1)^2} dt$  

これを元の式に代入すると

$\displaystyle \int{\sqrt{\frac{x+1}{x-1}}} dx = \int{\frac{-4t^2}{(t^2 - 1)^2}} dt$

この積分を行なうには部分分数分解を行なう．
$\displaystyle \frac{-4t^2}{(t^2 - 1)^{2}}$ $\displaystyle =$ $\displaystyle \frac{-4t^2}{(t+1)^{2}(t-1)^{2}}$  
  $\displaystyle =$ $\displaystyle \frac{A}{t+1} + \frac{B}{(t+1)^{2}} + \frac{C}{t-1} + \frac{D}{(t-1)^{2}}$  

分母を払うと

$\displaystyle -4t^2 = A(t+1)(t-1)^2 + B(t-1)^2 + C(t+1)^{2}(t-1) + D(t+1)^2$

$t = 1$とおくと

$\displaystyle -4 = 4D \Rightarrow D = -1$

$t = -1$とおくと

$\displaystyle -4 = 4B \Rightarrow B = -1$

$t^3$の係数を合わせると

$\displaystyle 0 = A + C$

$t^0$の係数を合わせると

$\displaystyle -4 = A + B - C + D \Rightarrow A - C = -2$

これより

$\displaystyle A = -1, C = 1が求まる．$

よって


$\displaystyle \int\frac{-4t^2}{(t^2 - 1)^2} dt$ $\displaystyle =$ $\displaystyle \int{\frac{-1}{t+1}} dt + \int{\frac{-1}{(t+1)^2}} dt$  
  $\displaystyle +$ $\displaystyle \int{\frac{1}{t-1}} dt + \int{\frac{-1}{(t-1)^2}} dt$  
  $\displaystyle =$ $\displaystyle -\log\vert t+1\vert + \frac{1}{t+1} + \log\vert t-1\vert + \frac{1}{t-1} + c$  
  $\displaystyle =$ $\displaystyle -\log\vert t+1\vert + \log\vert t-1\vert + \frac{2t}{t^2 - 1} + c$  
  $\displaystyle =$ $\displaystyle \log\vert\sqrt{\frac{x+1}{x-1}} + 1\vert - \log\vert\sqrt{\frac{x+1}{x-1}}\vert + \frac{(x-1)}{2}\sqrt{\frac{x+1}{x-1}} + c$  

2.

(a) $t = x^2 - 4$とおくと$dt = 2xdx$より，この問題は置換積分で求めることができる．


$\displaystyle \int{\frac{x}{\sqrt{x^2 - 4}}} dx$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int{\frac{dt}{\sqrt{t}}}$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}\int{t^{-1/2}} dt = \frac{1}{2}\cdot 2t^{1/2} + c$  
  $\displaystyle =$ $\displaystyle \sqrt{x^2 - 4} + c$  

(b) $t = \sqrt{4 - x^2}$とおくと $t^2 = 4 - x^2$より $2tdt = -2xdx$となり，全てを$t$と$dt$で表すと再び無理関数が登場してしまう．そこで，平方根の中が2乗の差であることに注意し，斜辺2，高さ$x$，底辺 $\sqrt{4 - x^2}$で角$t$の直角三角形を考える．すると $x = 2\sin{t}$より $dx = 2\cos{t}dt$．また，

$\displaystyle \sqrt{4 - x^2} = \sqrt{4 - 4\sin^{2}{t}} = \sqrt{4\cos^{2}{t}} = 2\cos{t}$

より


$\displaystyle \int{\frac{x^2}{\sqrt{4 - x^2}}} dx$ $\displaystyle =$ $\displaystyle \int{\frac{4\sin^{2}{t}}{2\cos{t}}2\cos{t}} dt = 4\int{\sin^{2}{t}} dt$  
  $\displaystyle =$ $\displaystyle 4\int{\frac{1 - \cos{2t}}{2}} dt = 2[t - \frac{\sin{2t}{2}}] + c$  
  $\displaystyle =$ $\displaystyle 2[t - \sin{t}\cos{t}] + c = 2[\sqrt{4 - x^2} - \frac{x}{2}\frac{\sqrt{4 - x^2}}{2} + c$  

(c) $t = e^{x}$とおくと $dt = e^{x} dx$となり，全てを$t$と$dt$で表すことができる． よって

$\displaystyle \int{\frac{e^{x}}{9 - e^{2x}}} dx$ $\displaystyle =$ $\displaystyle \int{\frac{1}{9 - t^2}} dt$  
  $\displaystyle =$ $\displaystyle \frac{1}{6}\int{\frac{1}{3 - t} + \frac{1}{3+t}} dt$  
  $\displaystyle =$ $\displaystyle \frac{1}{6}[-\log\vert 3-t\vert + \log\vert 3 + t\vert] +c = \frac{1}{6}\log\vert\frac{3+t}{3-t}\vert + c$  
  $\displaystyle =$ $\displaystyle \frac{1}{6}\log\vert\frac{3 + e^{x}}{3 - e^{x}}\vert + c$  

(d) $t = \sqrt{1 - x^2}$とおくと $t^2 = 1 - x^2$より $2tdt = -2xdx$となり，全てを$t$と$dt$で表すと再び無理関数が登場してしまう．そこで，平方根の中が2乗の差であることに注意し，斜辺1，高さ$x$，底辺 $\sqrt{1 - x^2}$で角$t$の直角三角形を考える．すると $x = \sin{t}$より $dx = \cos{t}dt$．また，

$\displaystyle \sqrt{1 - x^2} = \sqrt{1 - \sin^{2}{t}} = \sqrt{\cos^{2}{t}} = \cos{t}$

より


$\displaystyle \int{\frac{\sqrt{1 - x^2}}{x^4}} dx$ $\displaystyle =$ $\displaystyle \int{\frac{\cos{t}}{\sin^{4}{t}}\cos{t}} dt = \int{\frac{\cos^{2}{t}}{\sin^{4}{t}}} dt$  

ここで，全ての三角関数は有理関数に直せることに注意する．特にこの場合は分子，分母とも偶数乗であるので $u = \tan{t}$とおくと $t = \tan^{-1}{u}$より

$\displaystyle dt$ $\displaystyle =$ $\displaystyle \frac{1}{1 + u^2} du$  
$\displaystyle \sin{t}$ $\displaystyle =$ $\displaystyle \frac{u}{\sqrt{1 + u^2}}$  
$\displaystyle \cos{t}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{1 + u^2}}$  

となる．これより
$\displaystyle \int{\frac{\cos^{2}{t}}{\sin^{4}[t}} dt$ $\displaystyle =$ $\displaystyle \int{\frac{\frac{1}{1+u^2}}{\frac{u^4}{(1 + u^2)^{2}}}\frac{1}{1 + u^2}} du$  
  $\displaystyle =$ $\displaystyle \int{\frac{1}{u^{4}}} du = \int{u^{-4}} du$  
  $\displaystyle =$ $\displaystyle -\frac{1}{3}u^{-3} + c = -\frac{1}{3}\frac{1}{\tan^{3}{t}} + c$  
  $\displaystyle =$ $\displaystyle -\frac{1}{3}(\frac{\cos{t}}{\sin{t}})^{3} + c = -\frac{(1 - x^2)^{3/2}}{3x^3} + c$  

(e) $t = \sqrt{x^2 - a^2}$とおくと $t^2 = x^2 - a^2$より $2t dt = 2x dx$となり，全てを$t$と$dt$で表すと再び無理関数が登場してしまう．そこで，平方根の中が2乗の差であることに注意し，斜辺$x$，底辺$a$，高さ $\sqrt{x^2 - a^2}$で角$t$の直角三角形を考える．すると $x = a\sec{t}$より $dx = a\sec{t}\tan{t}dt$．また，

$\displaystyle \sqrt{x^2 - a^2} = \sqrt{a^{2}\sec^{2}{t} - a^2} = \sqrt{a^{2}\tan^{2}{t}} = a\tan{t}$

より


$\displaystyle \int{\frac{\sqrt{dx}}{x^2 \sqrt{x^2 - 4}}}$ $\displaystyle =$ $\displaystyle \int{\frac{a\sec{t}\tan{t}}{a^{2}\sec^{2}{t}a\tan{t}}} dt = \frac{1}{a^{2}}\int{\frac{1}{\sec{t}}} dt$  
  $\displaystyle =$ $\displaystyle \frac{1}{a^2}\int{\cos{t}} dt = \frac{1}{a^2}\sin{t} + c$  
  $\displaystyle =$ $\displaystyle \frac{\sqrt{x^2 - a^2}}{a^{2}x} + c$  

(f) $t = \sqrt{4 + e^{2x}}$とおくと $t^2 = 4 + e^{2x}$より $2tdt = 2e^{2x}dx$となり，全てを$t$と$dt$で表すと再び無理関数が登場してしまう．そこで，平方根の中が2乗の和であることに注意し，斜辺 $\sqrt{4 + e^{2x}}$，底辺$2$，高さ$e^{x}$で角$t$の直角三角形を考える．すると $e^{x} = 2\tan{t}$より $e^{x}dx = 2\sec^{2}{t}dt$．また，

$\displaystyle \sqrt{4 + e^{2x}} = \sqrt{4 + (2\tan{t})^{2}} = \sqrt{4\sec^{2}{t}} = 2\sec{t}$

より


$\displaystyle \int{\frac{\sqrt{dx}}{e^{x}\sqrt{4 + e^{2x}}}}$ $\displaystyle =$ $\displaystyle \int{\frac{1}{2\tan{t}2\sec{t}}\frac{2\sec^{2}{t}}{2\sec{t}}} dt$  
  $\displaystyle =$ $\displaystyle \frac{1}{4}\int{\frac{\sec{t}}{\tan^{2}{t}}} dt = \frac{1}{4}\int{\frac{\frac{1}{\cos{t}}}{\frac{\sin^{2}{t}}{\cos^{2}{t}}}} dt$  
  $\displaystyle =$ $\displaystyle \frac{1}{4}\int{\frac{\cos{t}}{\sin^{2}{t}}} dt = \frac{1}{4}\int{\frac{du}{u^{2}}}$  
  $\displaystyle =$ $\displaystyle -\frac{1}{4u} + c = -\frac{1}{4\sin{t}} + c = -\frac{\sqrt{4 + e^{2x}}}{4e^{x}} + c$  

(g) 平方根の中が2乗の和または差になるように平方完成を行なうと

$\displaystyle x^2 - 2x - 3 = (x-1)^2 - 4 = (x-1)^2 - 2^2$

斜辺$x-1$，底辺$2$，高さ $\sqrt{(x-1)^2 - 4}$で角$t$の直角三角形を考える．すると $x-1 = 2\sec{t}$より $dx = 2\sec{t}\tan{t}dt$．また，

$\displaystyle \sqrt{(x-1)^2 - 4} = \sqrt{4\sec^{2}{t} - 4} = \sqrt{4\tan^{2}{t}} = 2\tan{t}$

より
$\displaystyle \int{\frac{dx}{\sqrt{x^2 - 2x -3}}}$ $\displaystyle =$ $\displaystyle \int{\frac{2\sec{t}\tan{t}}{2\tan{t}}} dt$  
  $\displaystyle =$ $\displaystyle \int{\sec{t}} dt = \log\vert sec{t} + \tan{t}\vert + c$  
  $\displaystyle =$ $\displaystyle \log\vert\frac{x-1}{2} + \frac{sqrt{(x-1)^2 - 4}}{2}\vert + c$  

最後の部分で公式 $\int{\sec{t} dt} = \log\vert sec{t} + \tan{t}\vert + c$を用いたが，公式を用いなくとも次のように積分できる．
$\displaystyle \int{\sec{t}} dt$ $\displaystyle =$ $\displaystyle \int{\frac{1}{\cos{t}}} dt = \int{\frac{\cos{t}}{\cos^{2}{t}}} dt$  
  $\displaystyle =$ $\displaystyle \int{\frac{\cos{t}}{1 - \sin^{2}{t}}} dt$  
  $\displaystyle =$ $\displaystyle \int{\frac{du}{1 - u^{2}}} du = \frac{1}{2}\log\vert\frac{1+u}{1-u}\vert + c$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}\log\vert\frac{1+\cos{t}}{1 - \cos{t}}\vert + c$  

(h) 平方根の中が2乗の和または差になるように平方完成を行なうと

$\displaystyle 6x - x^2 = 9 - 9 + 6x - x^2 = 9 - (9 - 6x + x^2) = 3^2 - (3-x)^2$

斜辺$3$，底辺 $\sqrt{3^2 - (3-x)^2}$，高さ$3 - x$で角$t$の直角三角形を考える．すると $3 - x = 3\sin{t}$より $-dx = 3\cos{t}dt$．また，

$\displaystyle \sqrt{3^2 - (3-x)^2} = \sqrt{9 - 9\sin^{2}{t}} = \sqrt{9\cos^{2}{t}} = 3\cos{t}$

より
$\displaystyle \int{\frac{x}{\sqrt{3^2 - (3-x)^2}}} dx$ $\displaystyle =$ $\displaystyle \int{\frac{3 - 3\sin{t}}{3\cos{t}} (-3\cos{t})} dt$  
  $\displaystyle =$ $\displaystyle -\int{3 - 3\sin{t}} dt = -[3t + 3\cos{t}] + c$  
  $\displaystyle =$ $\displaystyle -3\sin^{-1}(\frac{3-x}{3}) - 3\frac{\sqrt{9 - (3-x)^2}}{3} + + c$  
  $\displaystyle =$ $\displaystyle -3\sin^{-1}(\frac{3-x}{3}) - \sqrt{6x - x^2} + c$  

(i) 平方根の中が2乗の和または差になるように平方完成を行なうと

$\displaystyle x^2 - 2x - 3 = (x-1)^2 -2^2$

斜辺$x-1$，底辺$2$，高さ $\sqrt{(x-1)^2 - 4}$で角$t$の直角三角形を考える．すると $x-1 = 2\sec{t}$より $dx = 2\sec{t}\tan{t}dt$．また，

$\displaystyle \sqrt{(x-1)^2 - 4} = \sqrt{4\sec^{2}{t} - 4} = \sqrt{4\tan^{t}} = 2\tan{t}$

より
$\displaystyle \int{\frac{x}{\sqrt{x^2 - 2x - 3}}} dx$ $\displaystyle =$ $\displaystyle \int{\frac{2\sec{t} + 1}{2\tan{t}} (2\sec{t}\tan{t})} dt$  
  $\displaystyle =$ $\displaystyle \int{(2\sec^{2}{t} + \sec{t})} dt$  
  $\displaystyle =$ $\displaystyle 2\tan{t} + \log\vert\sec{t} + \tan{t}\vert + c$  
  $\displaystyle =$ $\displaystyle 2(\frac{\sqrt{(x-1)^{2} - 4}}{2}) + \log\vert\frac{x-1}{2} + \frac{\sqrt{(x-1)^2 - 4}}{2} + c$  
  $\displaystyle =$ $\displaystyle \sqrt{x^2 - 2x - 3} + \log\vert\frac{x-1 + \sqrt{(x-1)^2 - 4}}{2}\vert + c$  

(j) 平方根の中が2乗の和または差になるように平方完成を行なうと

$\displaystyle 6x - x^2 - 8 = 9 - 9 + 6x - x^2 - 8 = 1 - (9 - 6x + x^2) = 1 - (3-x)^2 $

斜辺$1$，底辺 $\sqrt{1 - (3-x)^2}$，高さ$3 - x$で角$t$の直角三角形を考える．すると $3-x = \sin{t}$より $-dx = \cos{t}dt$．また，

$\displaystyle \sqrt{1 - (3-x)^2} = \sqrt{1 - \sin^{2}{t}} = \sqrt{\cos^{2}{t}} = \cos{t}$

より
$\displaystyle \int{\sqrt{6x - x^2 - 8}} dx$ $\displaystyle =$ $\displaystyle \int{\cos{t}(-\cos{t})} dt$  
  $\displaystyle =$ $\displaystyle -\int{\cos^{2}{t}} dt$  
  $\displaystyle =$ $\displaystyle -\int{\frac{1 + \cos{t}}{2}} dt = -[\frac{t}{2} + \frac{\sin{2t}}{4}] + c$  
  $\displaystyle =$ $\displaystyle -[\frac{t}{2} + \frac{\sin{t}\cos{t}}{2}] + c$  
  $\displaystyle =$ $\displaystyle -\frac{\sin^{-1}{(3-x)}}{2} - \frac{(3-x)\sqrt{1 - (3-x)^2}}{2} + c$  

(k) 平方根の中が2乗の和または差になるように平方完成を行なうと

$\displaystyle x^2 + 6x = (x+3)^2 - 9 $

斜辺$x+3$，底辺$3$，高さ $\sqrt{(x+3)^2 - 9}$で角$t$の直角三角形を考える．すると $x+3 = 3\sec{t}$より $dx = 3\sec{t}\tan{t}dt$．また，

$\displaystyle \sqrt{(x+3)^2 - 9} = \sqrt{9\sec^{2}{t} - 9} = \sqrt{9\tan^{2}{t}} = 3\tan{t}$

より
$\displaystyle \int{x\sqrt{x^2 + 6x}} dx$ $\displaystyle =$ $\displaystyle \int{(3\sec{t} - 3)3\tan{t}3\sec{t}\tan{t}} dt$  
  $\displaystyle =$ $\displaystyle 27\int{(\sec^{2}{t}\tan^{2}{t} - \sec{t}\tan^{2}{t}} dt$  
  $\displaystyle =$ $\displaystyle 27\int{(\frac{\sin^{2}{t}}{\cos^{4}{t}} - \frac{\sin^{2}{t}}{\cos^{3}{t}})} dt$  

ここで，全ての三角関数は有理関数に直せることに注意する． まず， $\int{\frac{\sin^{2}{t}}{\cos^{4}{t}}} dt$の積分を行なう． この場合は分子，分母とも偶数乗であるので $u = \tan{t}$とおくと $t = \tan^{-1}{u}$より

$\displaystyle dt$ $\displaystyle =$ $\displaystyle \frac{1}{1 + u^2} du$  
$\displaystyle \sin{t}$ $\displaystyle =$ $\displaystyle \frac{u}{\sqrt{1 + u^2}}$  
$\displaystyle \cos{t}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{1 + u^2}}$  

となる．これより
$\displaystyle \int{\frac{\sin^{2}{t}}{\cos^{4}{t}}} dt$ $\displaystyle =$ $\displaystyle \int{\frac{\frac{u^2}{1+u^2}}{\frac{1}{(1 + u^2)^{2}}}\frac{1}{1 + u^2}} du$  
  $\displaystyle =$ $\displaystyle \int{u^2} du = \frac{u^3}{3} + c$  
  $\displaystyle =$ $\displaystyle \frac{\tan^{3}{t}}{3} + c$  

次に， $\int{\frac{\sin^{2}{t}}{\cos^{3}{t}}} dt$の積分を行なう．この場合，分母に奇数乗を含んでいるので，分母と分子に$\cos{t}$をかける．すると
$\displaystyle \int{\frac{\sin^{2}{t}}{\cos^{3}{t}}} dt$ $\displaystyle =$ $\displaystyle \int{\frac{\sin^{2}{t}\cos{t}}{\cos^{4}{t}}} dt$  
  $\displaystyle =$ $\displaystyle \int{\frac{\sin^{2}{t}\cos{t}}{(1 - \sin^{2}{t})^{2}}} dt$  

ここで， $u = \sin{t}$とおくと $du = \cos{t} dt$となるので
$\displaystyle \int{\frac{\sin^{2}{t}\cos{t}}{(1 - \sin^{2}{t})^{2}}} dt$ $\displaystyle =$ $\displaystyle \int{\frac{u^2}{(1 - u^2)^{2}}} du = \int{\frac{u^2}{(1+u)^2 (1 -u)^2}} du$  

この積分を行なうには部分分数分解を行なう．

$\displaystyle \frac{u^2}{(1+u)^2 (1 -u)^2} = \frac{A}{1+u} + \frac{B}{(1+u)^{2}} + \frac{C}{1-u} + \frac{D}{(1-u)^{2}}$

分母を払うと

$\displaystyle u^2 = A(1+u)(1-u)^2 + B(1-u)^2 + C(1+u)^{2}(1-u) + D(1+u)^2$

$u=1$とおくと

$\displaystyle 1 = 4D \Rightarrow D = \frac{1}{4}$

$u=-1$とおくと

$\displaystyle 1 = 4B \Rightarrow B = \frac{1}{4}$

$t^3$の係数を合わせると

$\displaystyle 0 = A - C$

$t^0$の係数を合わせると

$\displaystyle 0 = A + B + C + D \Rightarrow A + C = \frac{-1}{2}$

これより

$\displaystyle A = \frac{-1}{4}, C = \frac{-1}{4}が求まる．$

よって


$\displaystyle \int\frac{u^2}{(1 - u^2)^2} du$ $\displaystyle =$ $\displaystyle \int{\frac{-1/4}{1+u}} du + \int{\frac{1/4}{(1+u)^2}} du$  
  $\displaystyle -$ $\displaystyle \int{\frac{1/4}{1-u}} du + \int{\frac{1/4}{(1-u)^2}} du$  
  $\displaystyle =$ $\displaystyle -\frac{1}{4}\log\vert 1+u\vert - \frac{1}{4}(\frac{1}{1+u}) + \frac{1}{4}(\log\vert 1-u\vert) + \frac{1}{4}(\frac{1}{1-u}) + c$  
  $\displaystyle =$ $\displaystyle \frac{1}{4}\log\vert\frac{1-u}{1+u}\vert - \frac{1}{4}(\frac{1}{1+u} - \frac{1}{1-u}) + c$  
  $\displaystyle =$ $\displaystyle \frac{1}{4}\log\vert\frac{1-u}{1+u}\vert + \frac{1}{4}(\frac{2u}{1-u^2}) + c$  
  $\displaystyle =$ $\displaystyle \frac{1}{4}\log\vert\frac{1-\sin{t}}{1+\sin{t}}\vert + \frac{1}{4}(\frac{2\sin{t}}{1-\sin^{2}{t}}) + c$  

したがって，

$\displaystyle \int{x\sqrt{x^2 + 6x}} dx$ $\displaystyle =$ $\displaystyle 27\int{\frac{\sin^{2}{t}}{\cos^{4}{t}} - \frac{\sin^{2}{t}}{\cos^{3}{t}}} dt$  
  $\displaystyle =$ $\displaystyle 27[\frac{\tan^{3}{t}}{3} + \frac{1}{4}\log\vert\frac{1-\sin{t}}{1+\sin{t}}\vert + \frac{1}{4}(\frac{2\sin{t}}{1-\sin^{2}{t}}) ]+ c$  
  $\displaystyle =$ $\displaystyle 27[\frac{1}{3}(\frac{\sqrt{x^2 + 6x}}{3})^{3} + \frac{1}{4}\log\v... ...(\frac{2\frac{\sqrt{x^2 + 6x}}{x+3}}{1-(\frac{\sqrt{x^2 + 6x}}{x+3})^{2}})] + c$  

next up previous index
Next: 3.7 解答 Up: 演習問題詳解 Previous: 3.5 解答   索引