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Homework 7
Section 8.1, #31: Compute $$\displaystyle\int \dfrac{\sin(x)}{\sqrt{\cos(x)}} \mathrm{d}x.$$ Solution: Let $u=\cos(x)$ so that $\mathrm{d}u=-\sin(x)\mathrm{d}x$ and hence $-\mathrm{d}u=\sin(x)\mathrm{d}x$. Therefore, compute $$\begin{array}{ll} \displaystyle\int \dfrac{\sin(x)}{\sqrt{\cos(x)}} \mathrm{d}x &= -\displaystyle\int u^{-\frac{1}{2}} \mathrm{d}u \\ &= -2 \sqrt{u} + C \\ &= -2 \sqrt{\cos(x)} + C. \end{array}$$ Section 8.2, #11: Compute $$\displaystyle\int x e^{-4x} \mathrm{d}x.$$ Solution: We will use integration by parts letting $u=x$ and $\mathrm{d}v=e^{-4x}\mathrm{d}x$. Hence $\mathrm{d}u=\mathrm{d}x$ and $v=-\dfrac{1}{4}e^{-4x}$. Therefore, compute $$\begin{array}{ll} \displaystyle\int x e^{-4x} \mathrm{d}x &= uv - \displaystyle\int v \mathrm{d}u \\ &= -\dfrac{x}{4} e^{-4x} - \displaystyle\int \left( -\dfrac{1}{4} e^{-4x} \right) \mathrm{d}x \\ &= -\dfrac{x}{4}e^{-4x}-\dfrac{1}{16} e^{-4x} + C. \end{array}$$ Note that WolframAlpha writes $-\dfrac{1}{16}e^{-4x}(4x+1)+C$, which is of course algebrically equivalent to the above.

Section 8.2, #26: Compute $$\displaystyle\int x^2 \cos(x) \mathrm{d}x.$$ Solution: Let $u=x^2$ and $\mathrm{d}v=\cos(x)\mathrm{d}x$ so that $\mathrm{d}u=2x\mathrm{d}x$ and $v=\sin(x)$. Now integration by parts tells us $$(*) \hspace{35pt} \begin{array}{ll} \displaystyle\int x^2 \cos(x) \mathrm{d}x &= uv - \displaystyle\int v \mathrm{d}u \\ &= x^2 \sin(x) - \displaystyle\int \sin(x) (2x) \mathrm{d}x \\ &= x^2 \sin(x) - 2\displaystyle\int x\sin(x) \mathrm{d}x. \end{array}$$ To evaluate the integral $\displaystyle\int x\sin(x)\mathrm{d}x$ we will again use integration by parts with $\tilde{u}=x$ and $\mathrm{d}\tilde{v}=\sin(x)\mathrm{d}x$ hence $\mathrm{d}\tilde{u}=\mathrm{d}x$ and $\tilde{v}=-\cos(x)$. Therefore use integration by parts to compute $$\begin{array}{ll} \displaystyle\int x \sin(x) \mathrm{d}x &= \tilde{u}\tilde{v} - \displaystyle\int \tilde{v} \mathrm{d}\tilde{u} \\ &= -x\cos(x) - \displaystyle\int (-\cos(x)) \mathrm{d}x \\ &=-x\cos(x) + \displaystyle\int \cos(x) \mathrm{d}x \\ &=-x\cos(x) + \sin(x) + C. \end{array}$$ We use this with our earlier equation $(*)$ to finally compute $$\begin{array}{ll} \displaystyle\int x^2 \cos(x) \mathrm{d}x &\stackrel{(*)}{=} x^2 \sin(x) - 2\displaystyle\int x \sin(x) \mathrm{d}x \\ &= x^2\sin(x) - 2 \left[ -x\cos(x) + \sin(x) \right] + C \\ &= (x^2-2)\sin(x) + 2x \cos(x) + C. \end{array}$$ Section 8.1, #45: Compute $$\displaystyle\int_0^1 e^x \sin(x) \mathrm{d}x.$$ Solution: We will approach this problem in two steps: first we will find the antiderivative (using integration by parts twice) and then we will use that antiderivative to find the given definite integral. Let $u=\sin(x)$ and $\mathrm{d}v=e^x \mathrm{d}x$ hence $\mathrm{d}u=\cos(x) \mathrm{d}x$ and $v=e^x$. Now integration by parts tells us that $$(*) \hspace{35pt} \begin{array}{ll} \displaystyle\int e^x \sin(x) \mathrm{d}x &= uv - \displaystyle\int v \mathrm{d}u \\ &=e^x \sin(x) - \displaystyle\int e^x \cos(x) \mathrm{d}x. \end{array}$$ To evaluate the integral $\displaystyle\int e^x \cos(x)$, we let $\tilde{u}=\cos(x)$ and $\mathrm{d}\tilde{v}=e^x \mathrm{d}x$ hence $\mathrm{d}\tilde{u}=-\sin(x)\mathrm{d}x$ and $\tilde{v}=e^x$. Therefore $$(**) \hspace{35pt} \begin{array}{ll} \displaystyle\int e^x \cos(x) \mathrm{d}x &= \tilde{u}\tilde{v} - \displaystyle\int \tilde{v} \mathrm{d}\tilde{u} \\ &= e^x \cos(x) - \displaystyle\int e^x (-\sin(x)) \mathrm{d}x \\ &= e^x \cos(x) + \displaystyle\int e^x \sin(x) \mathrm{d}x. \end{array}$$ Combining the equations $(*)$ and $(**)$ gives us the following formula: $$\displaystyle\int e^x \sin(x) \mathrm{d}x = e^x \sin(x) - e^x \cos(x) - \displaystyle\int e^x \sin(x) \mathrm{d}x.$$ Add $\displaystyle\int e^x \sin(x) \mathrm{d}x$ to both sides of this equation to obtain the equation $$2 \displaystyle\int e^x \sin(x) \mathrm{d}x = e^x \sin(x) - e^x \cos(x).$$ Therefore divide by $2$ and toss in the $+C$ to obtain $$\displaystyle\int e^x \sin(x) \mathrm{d}x = \dfrac{e^x \sin(x) - e^x \cos(x)}{2} + C.$$ Now we will use this antiderivative to compute $$\begin{array}{ll} \displaystyle\int_0^1 e^x \sin(x) \mathrm{d}x &= \dfrac{e^x \sin(x) - e^x \cos(x)}{2} \Bigg|_0^1 \\ &= \dfrac{e \sin(1)-e\cos(1)}{2} - \dfrac{1\sin(0) - 1\cos(0)}{2} \\ &= \dfrac{e \sin(1)-e\cos(1) + 1}{2}. \end{array}$$ Section 8.2, #44: Compute $$\displaystyle\int_0^1 x\arcsin(x^2) \mathrm{d}x.$$ Solution: First notice that the $u$-substitution $u=x^2$ hence $\dfrac{1}{2}\mathrm{d}u=x\mathrm{d}x$ yields $$(*) \hspace{35pt} \displaystyle\int_0^1 x\arcsin(x^2)\mathrm{d}x = \dfrac{1}{2} \displaystyle\int_0^1 \arcsin(u) \mathrm{d}u.$$ To finish, we will first find the antiderivative of $\arcsin(u)$ and then we will do the definite integral. Let $\tilde{u}=\arcsin(u)$ and $\mathrm{d}\tilde{v}=1 \mathrm{d}u$ hence $\mathrm{d}\tilde{u}=\dfrac{1}{\sqrt{1-u^2}} \mathrm{d}u$ and $\tilde{v}=u$. Therefore $$\begin{array}{ll} \displaystyle\int \arcsin(u) \mathrm{d}u &= \tilde{u}\tilde{v} - \displaystyle\int \tilde{v} \mathrm{d}\tilde{u} \\ &= u\arcsin(u) - \displaystyle\int \dfrac{u}{\sqrt{1-u^2}} \mathrm{d} u \\ &\stackrel{w=1-u^2}{=} u\arcsin(u) + \dfrac{1}{2} \displaystyle\int w^{-\frac{1}{2}} \mathrm{d}w \\ &=u\arcsin(u) + \sqrt{w}+C \\ &=u\arcsin(u) + \sqrt{1-u^2} +C. \end{array}$$ Therefore $$\begin{array}{ll} \displaystyle\int_0^1 x\arcsin(x^2) \mathrm{d}x &=\dfrac{1}{2} \displaystyle\int _0^1 \arcsin(u) \mathrm{d}u \\ &= \dfrac{1}{2} \left[ u\arcsin(u) + \sqrt{1-u^2} \right|_0^1 \\ &= \dfrac{1}{2} \left[ (1\arcsin(1)+\sqrt{0}) - (0+\sqrt{1}) \right] \\ &=\dfrac{\arcsin(1)-1}{2}\\ &\stackrel{\arcsin(1)=\frac{\pi}{2}}{=} \dfrac{\pi}{4} - \dfrac{1}{2}. \end{array}$$