置換積分
$$ \varphi (t) : C^1 級 \ (\varphi (t)は連続な導関数を持つ) $$
$$ f(x) : \varphi (t) の値域において連続 $$
$$ \implies \int_{a}^{b} f( \varphi(t) ) {\varphi}'(t) dt = \int_{\varphi (a)}^{\varphi (b)} f(x) dx $$
証明
$$ F(x) = \int f(x) dx \tag{1} $$
$$ G(t) = F(\varphi (t)) \tag{2} $$
とする。
$$ \begin{align} \frac{d}{dt} G (t) &= F’ (\varphi (t)) {\varphi}’ (t) \\ &= f( \varphi (t) ) {\varphi}’ (t) \quad \because (1) \end{align} $$
両辺を\(t\)で積分して、
$$ G(t) = \int ( \varphi (t) ) {\varphi}’ (t) dt \tag{3} $$
よって、
$$ \begin{align} \int_{a}^{b} f( \varphi(t) ) {\varphi}'(t) dt &= G(b) – G (a) \quad \because (3) \\ &= F(\varphi (b)) – F(\varphi (a)) \quad \because (2) \\ &= \int_{\varphi (a)}^{\varphi (b)} f(x) dx \quad \because (1) \end{align} $$
部分積分
$$ f, gが C^1 級のとき$$
$$ \int f'(x) g(x) dx = f(x) g(x) – \int f(x) g'(x) dx $$
$$ \int_{a}^{b} f'(x) g(x) dx = \left[ f(x) g(x) \right] _{a}^{b}- \int_{a}^{b} f(x) g'(x) dx $$
証明
$$ (f(x)g(x))’ = f'(x)g(x) + f(x) g'(x) $$
両辺積分して、
$$ \begin{align} f(x) g(x) &= \int \{f'(x)g(x) + f(x) g'(x) \} dx \\ &= \int f'(x)g(x) dx + \int f(x) g'(x) dx \end{align} $$
移項して、
$$ \int f'(x) g(x) dx = f(x) g(x) – \int f(x) g'(x) dx $$
例題
\(1 = (x)’\)を使う。
$$ \begin{align} \int \log x dx &= \int (x)’ \log x dx \\ &= x \log x – \int x \frac{1}{x} dx + C \\ &= x \log x – x + C \end{align} $$
部分積分で方程式を作る。
$$ \begin{align} \int \left( \frac{1}{x} \right) \log x dx &= \int (\log x)’ \log x dx \\ &= (\log x)^2 – \int \log x \left( \frac{1}{x} \right) dx \end{align} $$
$$ \therefore 2 \int \left( \frac{1}{x} \right) \log x dx = (\log x)^2 $$
$$ \int \left( \frac{1}{x} \right) \log x dx = \frac{(\log x)^2}{2} $$
部分積分を複数回行う。
$$ \begin{align} \int e^x \sin x dx &= \int (e^x)’ \sin x dx \\ &= e^x \sin x – \int e^x \cos x dx \\ &= e^x \sin x – \int (e^x)’ \cos x dx \\ &= e^x \sin x – \{ e^x \cos x + \int e^x \sin x dx \} \\ &= e^x (\sin x – \cos x) – \int e^x \sin x dx \end{align} $$
$$ \therefore \int e^x \sin x dx = \frac{e^x (\sin x – \cos x)}{2} $$
\( \int (f(x))^n dx \)を積分する。
$$ \begin{align} \int \sin^n x dx &= \int (\sin^{n-1} x ) (\sin x) dx \\ &= \int (\sin^{n-1} x ) (- \cos x)’ dx \\ &= (\sin^{n-1} x) (- \cos x) – \int (n-1) (\sin^{n-2} x ) (\cos x) (- \cos x) dx \\ &= – (\sin^{n-1} x)(\cos x) + (n-1) \int (\sin^{n-2} x)(1 – \sin^2 x) dx \\ &= – (\sin^{n-1} x)(\cos x) + (n-1) \int (\sin^{n-2} x) dx – (n-1) \int \sin^n x dx \end{align} $$
$$ n \int \sin^n x dx = – (\sin^{n-1} x)(\cos x) + (n-1) \int (\sin^{n-2} x) dx $$
$$ \int \sin^n x dx = – \frac{1}{n} (\sin^{n-1} x)(\cos x) + \frac{n-1}{n} \int (\sin^{n-2} x) dx $$
\(\int \sin^n x dx\)から\(\int \sin^{n-2} x dx\)へと指数が下がっているので、複数回適用すれば積分を終わらせることができる。
$$ \begin{align} \int \cos^n x dx &= \int (\cos^{n-1} x ) (\cos x) dx \\ &= \int (\cos^{n-1} x ) ( \sin x)’ dx \\ &= (\cos^{n-1} x) ( \sin x) – \int (n-1) (\cos^{n-2} x ) (- \sin x) ( \sin x) dx \\ &= (\cos^{n-1} x)(\sin x) + (n-1) \int (\cos^{n-2} x)(1 – \cos^2 x) dx \\ &= (\cos^{n-1} x)(\sin x) + (n-1) \int (\cos^{n-2} x) dx – (n-1) \int \cos^n x dx \end{align} $$
$$\int \cos^n x dx = \frac{1}{n}(\cos^{n-1} x)(\sin x) + \frac{n-1}{n} \int (\cos^{n-2} x) dx $$