-
$\begin{array}{cl}
& \dint_0^1 x^2e^{ax}dx \\
= & \dfrac{1}{a}\dint_0^1x^2d(e^{ax}) \\
= & \left[ \dfrac{1}{a}x^2e^{ax}\right]_0^1-\dfrac{1}{a}\dint_0^1 e^{ax}d(x^2) \\
= & \dfrac{e^a}{a} -\dfrac{2}{a}\dint_0^1 xe^{ax}dx \\
= & \dfrac{e^a}{a}-\dfrac{2}{a^2} \dint_0^1 xd(e^{ax}) \\
= & \dfrac{e^a}{a}-\left[ \dfrac{2}{a^2}xe^{ax}\right]_0^1+\dfrac{2}{a^2}\dint_0^1 e^{ax}dx \\
= & \dfrac{e^a}{a} -\dfrac{2e^a}{a^2}+\dfrac{2}{a^2}\left[ \dfrac{1}{a}e^{ax}\right]_0^1 \\
= & \dfrac{e^a}{a}-\dfrac{2e^a}{a^2}+\dfrac{2}{a^3}(e^a-1) \\
= & \dfrac{a^2e^a-2ae^a+2e^a-2}{a^3} \\
= & \dfrac{(a^2-2a+2)e^a-2}{a^3}
\end{array}$ - Let $u=\ln(1+x)$. Then we have
$\begin{array}{rcl}
du & = & \dfrac{1}{1+x} dx \\
e^udu & = & dx
\end{array}$When $x=0$, $u=0$.
When $x=e-1$, $u=1$.
Hence, we have
$\begin{array}{cl}
& \dint_0^{e-1} x\left(\ln(1+x)\right)^2 dx \\
= & \dint_0^1 (e^u-1)u^2e^u du \\
= & \dint_0^1 \left( u^2e^{2u} -u^2e^u\right) du \\
= & \dint_0^1 u^2e^{2u}du – \dint_0^1 u^2e^u du \\
= & \dfrac{(2^2-2(2)+2)e^2-2}{2^3}-\dfrac{(1^2-2(1)+2)e-2}{1^3} \text{ , by using (a).}\\
= & \dfrac{1}{4}e^2-\dfrac{1}{4}-e+2 \\
= & \dfrac{e^2}{4}-e+\dfrac{7}{4} \\
= & \dfrac{1}{4}(e^2-4e+7)
\end{array}$ - Let $u=(e-1)\cos x$. Then we have
$\begin{array}{rcl}
du & = & -(e-1)\sin x dx \\
\dfrac{-1}{e-1}du & = & \sin x dx
\end{array}$When $x=0$, $u=e-1$.
When $x=\dfrac{\pi}{2}$, $u=0$.
Hence, we have
$\begin{array}{cl}
& \dint_0^\frac{\pi}{2} \left(\ln(1+(e-1)\cos x)\right)^2\sin 2x dx \\
= & 2\dint_0^\frac{\pi}{2} \left(\ln(1+(e-1)\cos x)\right)^2\cos x\sin x dx \\
= & 2\dint_{e-1}^0 \left(\ln(1+u)\right)^2 \left(\dfrac{u}{e-1}\right)\left(\dfrac{-1}{e-1}\right) du \\
= & \dfrac{2}{(e-1)^2} \dint_0^{e-1}u(\ln(1+u))^2du \\
= & \dfrac{2}{(e-1)^2} \times\dfrac{1}{4}(e^2-4e+7) \text{ , by the result of (b).}\\
= & \dfrac{e^2-4e+7}{2(e-1)^2}
\end{array}$ - Let $u=x-\dfrac{\pi}{2}$. Then $du=dx$.
When $x=\dfrac{\pi}{2}$, $u=0$.
When $x=\pi$, $u=\dfrac{\pi}{2}$.
Hence, we have
$\begin{array}{cl}
& \dint_\frac{\pi}{2}^\pi \left(\ln(1+(e-1)\sin x)\right)^2\sin2x dx \\
= & \dint_0^\frac{\pi}{2} \left(\ln\left(1+(e-1)\sin\left(\dfrac{\pi}{2}+u\right)\right)\right)^2\sin2\left(u+\dfrac{\pi}{2}\right) du \\
= & \dint_0^\frac{\pi}{2} \left(\ln (1+(e-1)\cos u)\right)^2 \sin (\pi+2u) du \\
= & -\dint_0^\frac{\pi}{2} \left(\ln (1+(e-1)\cos u)\right)^2 \sin 2u du \\
= & \dfrac{-(e^2-4e+7)}{2(e-1)^2} \text{ , by the result of (c).}
\end{array}$
2023-M2-12
Ans: (b) $\dfrac{1}{4}(e^2-4e+7)$ (c) $\dfrac{e^2-4e+7}{2(e-1)^2}$ (d) $\dfrac{-e^2+4e-7}{2(e-1)^2}$
