３つの関数の積の積分

３つの関数の積の積分が３つの関数の積の導関数の公式の逆演算で解けるのではないかと思い試してみました。

\(\int_0^t(e^{t-s}cos\ s+\frac{(t^2-s^2)e^{t-s}sin s}{2})ds+2e^t+\frac{e^tt^2}{2}\)の解き方

\(\int_0^t(e^{t-s}cos s+\frac{(t^2-s^2)e^{t-s}sin s}{2})ds+2e^t+\frac{e^tt^2}{2}\)を解くのに必要な道具

1. ３つの関数の積の導関数の公式の逆演算
2. 部分積分

「３つの関数の積の導関数の公式」の逆演算

３つの関数の積の導関数の公式　\(y’=f’gh+fg’h+fgh’\)

「３つの関数の積の導関数の公式」の証明

積の導関数の公式より\(y’=(fg)’h+(fg)h’,\ y’=(f’g+fg’)h+(fg)h’,\ y’=f’gh+fg’h+fgh’\)

「３つの関数の積の導関数の公式」の逆演算

$$((f(x)g(x)h(x))’=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$$

両辺を積分して整理すると\(\int f(x)g(x)h'(x)dx=f(x)g(x)h(x)-\int f'(x)g(x)h(x)dx-\int f(x)g'(x)h(x)dx\)

\(\int h(x)dx=H(x)\)と置いて書き換えると\(\int f(x)g(x)h(x)dx=f(x)g(x)H(x)-\int f'(x)g(x)H(x)dx-\int f(x)g'(x)H(x)dx\)

部分積分

$$\int_{}{}f(x)g(x)dx=F(x)g(x)-\int_{}{}F(x)g'(x)dx$$

\(\int_0^t(e^{t-s}cos s+\frac{(t^2-s^2)e^{t-s}sin s}{2})ds+2e^t+\frac{e^tt^2}{2}\)の解き方

\(\int_{}{}e^{t-s}cos\ s\ ds\)の計算

$$\int_{}{}e^{t-s}sin\ s\ ds=-e^{t-s}cos\ s-\int_{}{}e^{t-s}cos\ s\ ds$$

$$\int_{}{}e^{t-s}cos\ s\ ds=e^{t-s}sin\ s-e^{t-s}cos\ s-\int_{}{}e^{t-s}cos\ s\ ds,\ 2\int_{}{}e^{t-s}cos\ s\ ds=e^{t-s}sin\ s-e^{t-s}cos\ s$$

$$\int_{}{}e^{t-s}cos\ s\ ds=\frac{1}{2}e^{t-s}sin\ s-\frac{1}{2}e^{t-s}cos\ s$$

\(\frac{t^2}{2}\int_{}{}e^{t-s}sin\ s\ ds\)の計算

$$\frac{t^2}{2}\int_{}{}e^{t-s}cos\ s\ ds=\frac{t^2}{2}\left(e^{t-s}sin\ s+\int_{}{}e^{t-s}sin\ s\ ds\right)$$

$$\frac{t^2}{2}\int_{}{}e^{t-s}sin\ s\ ds=\frac{t^2}{2}\left(-e^{t-s}cos\ s-e^{t-s}sin\ s-\int_{}{}e^{t-s}sin\ s\ ds\right)$$

$$t^2\int_{}{}e^{t-s}sin\ s\ ds=-\frac{t^2}{2}e^{t-s}cos\ s-\frac{t^2}{2}e^{t-s}sin\ s,\ \frac{t^2}{2}\int_{}{}e^{t-s}sin\ s\ ds=-\frac{t^2}{4}e^{t-s}cos\ s-\frac{t^2}{4}e^{t-s}sin\ s$$

\(-\frac{1}{2}\int_{}{}s^2e^{t-s}sin\ s\ ds\)の計算

$$-\frac{1}{2}\int_{}{}s^2e^{t-s}cos\ s\ ds=-\frac{1}{2}\left(s^2e^{t-s}sin\ s-2\int_{}{}se^{t-s}sin\ s\ ds+\int_{}{}s^2e^{t-s}sin\ s\ ds\right)$$

$$-\frac{1}{2}\int_{}{}s^2e^{t-s}sin\ s\ ds=-\frac{1}{2}\left(-s^2e^{t-s}cos\ s+2\int_{}{}se^{t-s}cos\ s\ ds-\int_{}{}s^2e^{t-s}cos\ s\ ds\right)$$

$$-\frac{1}{2}\int_{}{}s^2e^{t-s}sin\ s\ ds$$

$$=-\frac{1}{2}\left(-s^2e^{t-s}cos\ s+2\int_{}{}se^{t-s}cos\ s\ ds-s^2e^{t-s}sin\ s+2\int_{}{}se^{t-s}sin\ s\ ds-\int_{}{}s^2e^{t-s}sin\ s\ ds\right)$$

$$-\int_{}{}s^2e^{t-s}sin\ s\ ds=-\frac{1}{2}\left(-s^2e^{t-s}cos\ s+2\int_{}{}se^{t-s}cos\ s\ ds-s^2e^{t-s}sin\ s+2\int_{}{}se^{t-s}sin\ s\ ds\right)$$

$$-\frac{1}{2}\int_{}{}s^2e^{t-s}sin\ s\ ds=-\frac{1}{4}\left(-s^2e^{t-s}cos\ s+2\int_{}{}se^{t-s}cos\ s\ ds-s^2e^{t-s}sin\ s+2\int_{}{}se^{t-s}sin\ s\ ds\right)$$

\(2\int{}{}se^{t-s}cos\ s\ ds\)の計算

$$2\int{}{}se^{t-s}sin\ s\ ds=2\left(-se^{t-s}cos\ s+\int_{}{}e^{t-s}cos\ s \ sd-\int_{}{}se^{t-s}cos\ s\ ds\right)$$

$$2\int{}{}se^{t-s}cos\ s\ ds=2\left(se^{t-s}sin\ s-\int_{}{}e^{t-s}sin\ s \ ds-se^{t-s}cos\ s+\int_{}{}e^{t-s}cos\ s\ ds-\int_{}{}se^{t-s}cos\ s\ ds\right)$$

$$4\int{}{}se^{t-s}cos\ s\ ds=2\left(se^{t-s}sin\ s-\int_{}{}e^{t-s}sin\ s \ ds-se^{t-s}cos\ s+\int_{}{}e^{t-s}cos\ s\ ds\right)$$

$$2\int{}{}se^{t-s}cos\ s\ ds=se^{t-s}sin\ s-\int_{}{}e^{t-s}sin\ s \ ds-se^{t-s}cos\ s+\int_{}{}e^{t-s}cos\ s\ ds$$

\(\int_{}{}e^{t-s}sin\ s\ ds\)の計算

$$\int_{}{}e^{t-s}cos\ s\ ds=e^{t-s}sin\ s+\int_{}{}e^{t-s}sin\ s\ ds$$

$$\int_{}{}e^{t-s}sin\ s\ ds=-e^{t-s}cos\ s-e^{t-s}sin\ s\ ds-\int_{}{}e^{t-s}sin\ s\ ds$$

$$2\int_{}{}e^{t-s}sin\ s\ ds=-e^{t-s}cos\ s-e^{t-s}sin\ s,\ \int_{}{}e^{t-s}sin\ s\ ds=-\frac{1}{2}e^{t-s}cos\ s-\frac{1}{2}e^{t-s}sin\ s$$

\(\int_{}{}e^{t-s}cos\ s\ ds\)の計算

$$\int_{}{}e^{t-s}cos\ s\ ds=e^{t-s}sin\ s+\int_{}{}e^{t-s}sin\ s\ ds,\ \int_{}{}e^{t-s}cos\ s\ ds=e^{t-s}sin\ s-e^{t-s}cos\ s-\int_{}{}e^{t-s}cos\ s\ ds$$

$$2\int_{}{}e^{t-s}cos\ s\ ds=e^{t-s}sin\ s-e^{t-s}cos\ s,\ \int_{}{}e^{t-s}cos\ s\ ds=\frac{1}{2}e^{t-s}sin\ s-\frac{1}{2}e^{t-s}cos\ s$$

\(2\int{}{}se^{t-s}cos\ s\ ds\)の計算の続き

$$2\int{}{}se^{t-s}cos\ s\ ds=se^{t-s}sin\ s+\frac{1}{2}e^{t-s}cos\ s+\frac{1}{2}e^{t-s}sin\ s-se^{t-s}cos\ s+\frac{1}{2}e^{t-s}sin\ s-\frac{1}{2}e^{t-s}cos\ s$$

$$2\int{}{}se^{t-s}cos\ s\ ds=-se^{t-s}cos\ s+se^{t-s}sin\ s+e^{t-s}sin\ s$$

\(2\int{}{}se^{t-s}sin\ s\ ds\)の計算

$$2\int{}{}se^{t-s}cos\ s\ ds=2\left(se^{t-s}sin\ s-\int_{}{}e^{t-s}sin\ s \ ds+\int_{}{}se^{t-s}sin\ s\ ds\right)$$

$$2\int{}{}se^{t-s}sin\ s\ ds=2\left(-se^{t-s}cos\ s+\int_{}{}e^{t-s}cos\ s \ ds-\int_{}{}se^{t-s}cos\ s\ ds\right)$$

$$2\int{}{}se^{t-s}sin\ s\ ds=2\left(-se^{t-s}cos\ s+\int_{}{}e^{t-s}cos\ s \ ds-se^{t-s}sin\ s+\int_{}{}e^{t-s}sin\ s \ ds-\int_{}{}se^{t-s}sin\ s\ ds\right)$$

$$4\int{}{}se^{t-s}sin\ s\ ds=2\left(-se^{t-s}cos\ s+\int_{}{}e^{t-s}cos\ s \ ds-se^{t-s}sin\ s+\int_{}{}e^{t-s}sin\ s \ ds\right)$$

$$2\int{}{}se^{t-s}sin\ s\ ds=-se^{t-s}cos\ s+\int_{}{}e^{t-s}cos\ s \ ds-se^{t-s}sin\ s+\int_{}{}e^{t-s}sin\ s \ ds$$

$$=-se^{t-s}cos\ s+\frac{1}{2}e^{t-s}sin\ s-\frac{1}{2}e^{t-s}cos\ s-se^{t-s}sin\ s-\frac{1}{2}e^{t-s}cos\ s-\frac{1}{2}e^{t-s}sin\ s$$

$$2\int{}{}se^{t-s}sin\ s\ ds=-se^{t-s}cos\ s-se^{t-s}sin\ s-e^{t-s}cos\ s$$

\(-\frac{1}{2}\int_{}{}s^2e^{t-s}sin\ s\ ds\)の計算の続き

$$-\frac{1}{2}\int_{}{}s^2e^{t-s}sin\ s\ ds=-\frac{1}{4}(-s^2e^{t-s}cos\ s-se^{t-s}cos\ s+se^{t-s}sin\ s$$

$$+e^{t-s}sin\ s-s^2e^{t-s}sin\ s-se^{t-s}cos\ s-se^{t-s}sin\ s-e^{t-s}cos\ s)$$

$$=-\frac{1}{4}(-s^2e^{t-s}cos\ s-2se^{t-s}cos\ s+e^{t-s}sin\ s-s^2e^{t-s}sin\ s-e^{t-s}cos\ s)$$

\(\int_0^te^{t-s}cos\ s\ ds\)の計算

$$\int_0^te^{t-s}cos\ s\ ds=\left[\frac{1}{2}e^{t-s}sin\ s-\frac{1}{2}e^{t-s}cos\ s\right]^t_0=\frac{1}{2}sin\ t-\frac{1}{2}cos\ t+\frac{1}{2}e^t$$

\(\frac{t^2}{2}\int_0^te^{t-s}sin\ s\ ds\)の計算

$$\frac{t^2}{2}\int_0^te^{t-s}sin\ s\ ds=\left[-\frac{t^2}{4}e^{t-s}cos\ s-\frac{t^2}{4}e^{t-s}sin\ s\right]^t_0=-\frac{t^2}{4}cos\ t-\frac{t^2}{4}sin\ t+\frac{t^2}{4}e^t$$

\(-\frac{1}{2}\int_0^ts^2e^{t-s}sin\ s\ ds\)の計算

$$-\frac{1}{2}\int_0^ts^2e^{t-s}sin\ s\ ds=\left[\frac{1}{4}s^2e^{t-s}cos\ s+\frac{1}{2}se^{t-s}cos\ s-\frac{1}{4}e^{t-s}sin\ s-\frac{1}{4}s^2e^{t-s}sin\ s+\frac{1}{4}e^{t-s}cos\ s\right]^t_0$$

$$=\frac{t^2}{4}cos\ t+\frac{t}{2}cos\ t-\frac{1}{4}sin\ t+\frac{t^2}{4}sin\ t+\frac{t}{4}cos\ t-\frac{1}{4}e^t$$

\(\int_0^t(e^{t-s}cos\ s+\frac{(t^2-s^2)e^{t-s}sin s}{2})ds+2e^t+\frac{e^tt^2}{2}\)の計算

$$\int_0^t(e^{t-s}cos\ s+\frac{(t^2-s^2)e^{t-s}sin s}{2})ds+2e^t+\frac{e^tt^2}{2}=\frac{1}{4}sin\ t-\frac{1}{4}cos\ t+\frac{9}{4}e^t+\frac{3}{4}t^2e^t+\frac{2}{4}tcos\ t$$

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