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Section 8.2 #11: $\displaystyle\int x e^{-4x} \mathrm{d}x$
Solution: Let $u=x$ and $\mathrm{d}v=e^{-4x}$. Then $\mathrm{d}u=\mathrm{d}x$ and $v = -\dfrac{1}{4} e^{-4x}$. Therefore using integration by parts we compute $$\begin{array}{ll} \displaystyle\int xe^{-4x} \mathrm{d}x &= -\dfrac{1}{4}xe^{-4x} + \dfrac{1}{4} \displaystyle\int e^{-4x} \mathrm{d}x \\ &= -\dfrac{1}{4}xe^{-4x} - \dfrac{1}{16} e^{-4x} + C. \end{array}$$

Section 8.2 #13: $\displaystyle\int x^3 e^x \mathrm{d}x$
Solution: Let $u_1=x^3$ and $\mathrm{d}v_1=e^x$. Then $\mathrm{d}u_1 = 3x^2 \mathrm{d}x$ and $v_1 = e^x$. Therefore one application of integration by parts yields $$(*) \hspace{35pt} \displaystyle\int x^3 e^x \mathrm{d}x = x^3 e^x - 3\displaystyle\int x^2 e^x \mathrm{d}x.$$ Now we will compute that integral. Let $u_2=x^2$ and $\mathrm{d}v_2=e^x$. Then $\mathrm{d}u_2 = 2x \mathrm{d}x$ and $v_2=e^x$ yielding $$(**) \hspace{35pt} \displaystyle\int x^2 e^x \mathrm{d}x = x^2 e^x - 2 \displaystyle\int x e^x \mathrm{d}x.$$ Now we will compute that integral. Let $u_3=x$ and $\mathrm{d}v_3=e^x$. Then $\mathrm{d}u_3 = \mathrm{d}x$ and $v_3=e^x$ yielding $$(***) \hspace{35pt} \displaystyle\int xe^x \mathrm{d}x = xe^x - \displaystyle\int e^x \mathrm{d}x = xe^x - e^x + C.$$ Combining $(*)$, $(**)$, and $(***)$ allows us to compute $$\begin{array}{ll} \displaystyle\int x^3 e^x \mathrm{d}x &= x^3e^x - 3 \left( \displaystyle\int x^2 e^x \mathrm{d}x \right) \\ &= x^3 e^x - 3 \left( x^2e^x - 2 \left( \displaystyle\int xe^x \mathrm{d}x \right) \right) \\ &= x^3 e^x - 3 \left( x^2 e^x - 2 \left( xe^x -e^x \right) \right) + C \\ &= x^3e^x - 3x^2e^x + 6xe^x - 6e^x + C. \end{array}$$

Section 8.2 #26: $\displaystyle\int x^2 \cos(x) \mathrm{d}x$
Solution: Let $u_1=x^2$ and $\mathrm{d}v_1=\cos(x)$. Then $\mathrm{d}u_1=2x\mathrm{d}x$ and $v_1=\sin(x)$ and so $$(*) \hspace{35pt} \displaystyle\int x^2 \cos(x) \mathrm{d}x = x^2 \sin(x) - 2\displaystyle\int x\sin(x).$$ Now we compute that integral. Let $u_2=x$ and $\mathrm{d}v_2=\sin(x)$. Then $\mathrm{d}u_2=\mathrm{d}x$ and $v_2=-\cos(x)$ and so $$(**) \hspace{35pt} \displaystyle\int x \sin(x) \mathrm{d}x = -x\cos(x) + \displaystyle\int \cos(x) \mathrm{d}x=-x\cos(x) + \sin(x) + C.$$ Combining $(*)$ and $(**)$ allows us to compute $$\begin{array}{ll} \displaystyle\int x^2 \cos(x) \mathrm{d}x &= x^2 \sin(x) - 2 \left( \displaystyle\int x \sin(x) \mathrm{d}x \right) \\ &= x^2 \sin(x) - 2 \left( -x \cos(x) + \sin(x) \right) + C \\ &= x^2\sin(x) - 2\sin(x) + 2x \cos(x) + C. \end{array}$$ Section 8.2 #45: $\displaystyle\int_0^1 e^x \sin(x) \mathrm{d}x$
Solution: First we will find the antiderivative of $e^x \sin(x)$. To do this, first let $u_1=e^x$ and $\mathrm{d}v_1=\sin(x)$. Then $\mathrm{d}u_1=e^x \mathrm{d}x$ and $v_1=-\cos(x)$ and so $$(*) \hspace{35pt} \displaystyle\int e^x \sin(x) \mathrm{d}x = -e^x \cos(x) + \displaystyle\int e^x \cos(x) \mathrm{d}x.$$ Now we compute that integral. Let $u_2=e^x$ and $\mathrm{d}v_2=\cos(x) \mathrm{d}x$. Then $\mathrm{d}u_2=e^x \mathrm{d}x$ and $v_2 = \sin(x)$ so we compute $$(**) \hspace{35pt} \displaystyle\int e^x \cos(x) \mathrm{d}x = e^x \sin(x) - \displaystyle\int e^x \sin(x) \mathrm{d}x.$$ Therefore we may use $(*)$ and $(**)$ to compute $$\begin{array}{ll} \displaystyle\int e^x \sin(x) \mathrm{d}x &= -e^x \cos(x) + \displaystyle\int e^x \cos(x) \mathrm{d}x \\ &= -e^x \cos(x) + e^x \sin(x) - \displaystyle\int e^x \sin(x) \mathrm{d}x. \end{array}$$ Therefore add $\displaystyle\int e^x \sin(x) \mathrm{d}x$ to both sides to get $$2 \displaystyle\int e^x \sin(x) \mathrm{d}x = e^x \sin(x) - e^x \cos(x).$$ Dividing by $2$ yields the antiderivative $$\displaystyle\int e^x \sin(x) \mathrm{d}x = \dfrac{e^x \sin(x) - e^x \cos(x)}{2}.$$ Now we apply this antiderivative to find the definite integral in question: compute $$\begin{array}{ll} \displaystyle\int_0^1 e^x \sin(x) \mathrm{d}x &= \left. \dfrac{e^x \sin(x) - e^x \cos(x)}{2} \right|_0^1 \\ &= \dfrac{e \sin(1) - e\cos(1)}{2} - \dfrac{0 -1}{2} \\ &=\dfrac{e\sin(1) -e \cos(1) + 1}{2}. \end{array}$$