部分積分

$\displaystyle \int f(x)g'(x)~dx = f(x)g(x) - \int f'(x)g(x)~dx$

を利用しましょう。

(1) $\sin x = (-\cos x)'$ であるから

$\begin{eqnarray*} \int x\sin x~dx & = & \int x(-\cos x)' ~dx\\[0.5em] & = & x(-\cos x) - \int (x)'(-\cos x)~dx\\[0.5em] & = & -x\cos x + \int \cos x ~dx\\[0.5em] & = & -x\cos x + \sin x + C \end{eqnarray*}$

(2) $\cos 2x = \left(\dfrac{1}{2}\sin 2x \right)'$ であるから

$\begin{eqnarray*} \int x^2\cos 2x ~dx & = & \int x^2\left(\dfrac{1}{2}\sin 2x \right)' ~dx\\[1em] & = & x^2\left( \dfrac{1}{2}\sin 2x \right) - \int (x^2)'\left( \dfrac{1}{2}\sin 2x \right)~dx\\[1em] & = & \dfrac{1}{2}x^2\sin 2x - \int x\sin 2x ~dx \end{eqnarray*}$

ここで $\sin 2x = \left(-\dfrac{1}{2}\cos 2x \right)'$ であるから

$\begin{eqnarray*} \int x\sin 2x~dx & = & \int x\left(-\dfrac{1}{2}\cos 2x \right)' ~dx\\[1em] & = & x\left(-\dfrac{1}{2}\cos 2x \right) - \int (x)'\left(-\dfrac{1}{2}\cos 2x \right)~dx\\[1em] & = & -\dfrac{1}{2}x\cos 2x + \dfrac{1}{2}\int \cos 2x ~dx\\[1em] & = & -\dfrac{1}{2}x\cos 2x + \dfrac{1}{4}\sin 2x + C \end{eqnarray*}$

よって

$\begin{eqnarray*}\int x^2\cos 2x ~dx & = & \dfrac{1}{2}x^2\sin 2x - \int x\sin 2x ~dx\\[0.5em] & = & \dfrac{1}{2}x^2\sin 2x + \dfrac{1}{2}x\cos 2x - \dfrac{1}{4}\sin 2x + C\\[0.5em] & = & \dfrac{1}{4}\left( (2x^2-1)\sin 2x + 2x\cos 2x \right) +C \end{eqnarray*}$

(3) $1 = (x)'$ であるから

$\begin{eqnarray*} \int \log x~dx & = & \int (x)'\log x ~dx\\[0.5em] & = & x\log x - \int x(\log x)'~dx\\[0.5em] & = & x\log x - \int ~dx\\[0.5em] & = & x\log x - x + C \end{eqnarray*}$

(1)
部分積分を繰り返すと

$\begin{eqnarray*}\int_{-\pi}^{\pi} e^x \cos x ~dx & = & \left[e^x \sin x\right]_{-\pi}^{\pi} - \int_{-\pi}^{\pi} e^x\sin x~dx\\[0.5em] & = & 0-\left( \left[ -e^x \cos x\right]_{-\pi}^{\pi} - \int_{-\pi}^{\pi} -e^x\cos x~dx\right)\\[0.5em] & = & -( e^{\pi} - e^{-\pi}) - \int_{-\pi}^{\pi} e^x \sin x ~ dx \end{eqnarray*}$

$\displaystyle I = \int_{-\pi}^{\pi} e^x \cos x ~dx$ とすれば

$I = -(e^{\pi} - e^{-\pi}) - I$

よって $\displaystyle \int_{-\pi}^{\pi} e^x \cos x ~dx = \dfrac{e^{-\pi} - e^{\pi}}{2}$ となります。

(2)
部分積分を繰り返すと

$\begin{eqnarray*} \int_{-2}^{2}x^2 e^{-x}\ dx &=& \left[x^2 (-e^{-x})\right]_{-2}^2-\int_{-2}^{2} 2x(-e^{-x})\ dx \\[1em] &=& -4e^{-2}+4e^2+2\int_{-2}^{2}xe^{-x}\ dx\\[1em] & = & -4e^{-2}+4e^2+2 \left( \left[ x(-e^{-x}) \right]_{-2}^2-\int_{-2}^{2}(-e^{-x})\ dx \right)\\[1em] & = & -4e^{-2}+4e^2+2\left( -2e^{-2}-2e^2+\int_{-2}^{2}e^{-x}\ dx \right)\\[1em] & = & -4e^{-2}+4e^2 -4e^{-2} - 4e^2 + 2\int_{-2}^2e^{-x} ~dx\\[1em] & = & -8e^{-2} + 2\left[ - e^{-x} \right]_{-2}^2\\[1em] & = & -8e^{-2} - 2\left( e^{-2} - e^2 \right) = 2e^2 - 10e^{-2} \end{eqnarray*}$