横浜国立大学(理系) 2026年 問題1
$次の問いに答えよ。$
$(1)\ \ 関数\ f(x)=e^x\sin x,\ \ g(x)=e^x\cos x \ \ をそれぞれ微分せよ。$
\[(2)\ \ 次の定積分を求めよ。\quad \int_0^{\pi} xe^x\sin x dx\]
(1)
$f'(x)=e^x\sin x+e^x\cos x \hspace{5em}①$
$g'(x)=e^x\cos x-e^x\sin x \hspace{5em}②$
(2)
$①+② より \quad f'(x)+g'(x)=2e^x\cos x \qquad e^x\cos x=\dfrac{1}{2}\big(f'(x)+g'(x)\big)$
$①-② より \quad f'(x)-g'(x)=2e^x\sin x \qquad e^x\sin x=\dfrac{1}{2}\big(f'(x)-g'(x)\big)$
$よって \ \ C\ を積分定数として$
\[\int e^x\cos x dx=\int \dfrac{1}{2}\big(f'(x)+g'(x)\big)dx=\dfrac{1}{2}\big(f(x)+g(x)\big)+C=\dfrac{1}{2}\big(e^x\sin x + e^x\cos x\big) +C \] \[\int e^x\sin x dx=\int \dfrac{1}{2}\big(f'(x)-g'(x)\big)dx=\dfrac{1}{2}\big(f(x)-g(x)\big)+C=\dfrac{1}{2}\big(e^x\sin x - e^x\cos x\big) +C \]
\begin{eqnarray*} I &=&\int_0^{\pi} xe^x\sin x dx\\ \\ &=&\big[x \times \dfrac{1}{2}\big(e^x\sin x - e^x\cos x\big)\big]_0^{\pi} -\int_0^{\pi} \dfrac{1}{2}\big(e^x\sin x - e^x\cos x\big)dx\\ \\ &=&\dfrac{\pi}{2} e^{\pi} -\dfrac{1}{2}\int_0^{\pi} e^x\sin x dx + \dfrac{1}{2}\int_0^{\pi} e^x\cos x dx\\ \\ &=&\dfrac{\pi}{2} e^{\pi} -\dfrac{1}{4}\big[e^x\sin x - e^x\cos x \big]_0^{\pi} + \dfrac{1}{4}\big[e^x\sin x +e^x\cos x\big]_0^{\pi}\\ \\ &=&\dfrac{\pi}{2} e^{\pi} -\dfrac{1}{4}\big(e^{\pi}+1\big)+\dfrac{\pi}{4}\big(-e^{\pi}-1\big)\\ \\ &=&\dfrac{\pi}{2} e^{\pi} -\dfrac{1}{2}e^{\pi}-\dfrac{1}{2}\\ \\ &=&\dfrac{\pi-1}{2} e^{\pi} -\dfrac{1}{2} \end{eqnarray*}
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