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
- ３つの関数の積の導関数の公式の逆演算
- 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
$$\int_0^t(e^{t-s}cos\ s+\frac{(t^2-s^2)e^{t-s}sin s}{2})ds+2e^t+\frac{e^tt^2}{2}=\frac{1}{4}sin\ t-\frac{1}{4}cos\ t+\frac{9}{4}e^t+\frac{3}{4}t^2e^t+\frac{2}{4}tcos\ t$$
