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Section 5.6 #40: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t^2).$$ Solution: Since $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t)=\dfrac{1}{\sqrt{1-x^2}},$$ we, using the chain rule, compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t^2) &= \dfrac{1}{\sqrt{1-(t^2)^2}} \dfrac{\mathrm{d}}{\mathrm{d}t} t^2 \\ &= \dfrac{2t}{\sqrt{1-t^4}}. \blacksquare \end{array}$$

Section 5.6 #50: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \log(t^2+4) - \dfrac{1}{2} \arctan \left( \dfrac{t}{2} \right) \right].$$ Solution: Since $$\dfrac{\mathrm{d}}{\mathrm{d}t} \log(t) = \dfrac{1}{t}$$ and $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arctan(t) = \dfrac{1}{1+t^2},$$ we use the chain rule to compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \log(t^2+4) - \dfrac{1}{2} \arctan \left( \dfrac{t}{2} \right) \right] &= \dfrac{1}{t^2+4} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ t^2+4 \right] - \left( \dfrac{1}{2} \right) \dfrac{1}{1+\left(\frac{t}{2}\right)^2} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \dfrac{t}{2} \right] \\ &= \dfrac{2t}{t^2+4} -\dfrac{1}{4(1+\frac{t^2}{4})} \\ &= \dfrac{2t}{t^2+4} - \dfrac{1}{4+t^2}. \\ &= \dfrac{2t-1}{t^2+4}. \blacksquare \end{array}$$

Section 5.7 #17: Calculate $$\displaystyle\int \dfrac{x-3}{x^2+1} \mathrm{d}x.$$ Solution: Since $$\displaystyle\int f(x) + g(x) \mathrm{d}x = \displaystyle\int f(x) \mathrm{d}x + \displaystyle\int g(x) \mathrm{d}x,$$ we compute $$\begin{array}{ll} \displaystyle\int \dfrac{x-3}{x^2+1} \mathrm{d}x &= \displaystyle\int \dfrac{x}{x^2+1} \mathrm{d}x - 3 \displaystyle\int \dfrac{1}{x^2+1} \mathrm{d}x \\ &\stackrel{u=x^2+1, \frac{1}{2}\mathrm{d}u= x\mathrm{d}x}{=} \dfrac{1}{2} \displaystyle\int \dfrac{1}{u} \mathrm{d}u - 3 \arctan(x) + C \\ &= \dfrac{1}{2} \log(u) - 3 \arctan(x) + C \\ &= \dfrac{1}{2} \log(x^2+1) - 3\arctan(x) + C. \blacksquare \end{array}$$

Section 5.7 #28: Calculate $$\displaystyle\int_{\ln(2)}^{\ln(4)} \dfrac{e^{-x}}{\sqrt{1-e^{-2x}}} \mathrm{d}x.$$ Solution: Note that $$\sqrt{1-e^{-2x}} = \sqrt{1-(e^{-x})^2}.$$ Therefore calculate $$\begin{array}{ll} \displaystyle\int_{\ln(2)}^{\ln(4)} \dfrac{e^{-x}}{\sqrt{1-e^{-2x}}} \mathrm{d}x &\stackrel{u=e^{-x},-\mathrm{d}u=e^{-x}\mathrm{d}x}{=} -\displaystyle\int_{\frac{1}{2}}^{\frac{1}{4}} \dfrac{1}{\sqrt{1-u^2}} \mathrm{d}u \\ &= - \arcsin(u) \Bigg|_{\frac{1}{2}}^{\frac{1}{4}} \\ &=-\left[ \arcsin\left( \dfrac{1}{4} \right) - \arcsin \left( \dfrac{1}{2} \right) \right] \\ &= \arcsin \left( \dfrac{1}{2} \right) - \arcsin \left( \dfrac{1}{4} \right) \\ &\approx 0.2709. \blacksquare \end{array}$$

Section 5.7 #29: Calculate $$\displaystyle\int_{\frac{\pi}{2}}^{\pi} \dfrac{\sin(x)}{1+\cos^2(x)} \mathrm{d}x.$$ Solution: Let $u=\cos(x)$ so that $-\mathrm{d}u=\sin(x) \mathrm{d}x$. Now if $x=\dfrac{\pi}{2}$ then $u=\cos\left(\dfrac{\pi}{2}\right)=0$ and if $x=\pi$ then $u=cos(\pi)=-1$. Now calculate $$\begin{array}{ll} \displaystyle\int_{\frac{\pi}{2}}^{\pi} \dfrac{\sin(x)}{1+\cos^2(x)} \mathrm{d}x &= \displaystyle\int_{0}^{-1} \dfrac{1}{1+u^2} \mathrm{d}u \\ &= \arctan(u) \Bigg|_{0}^{-1} \\ &= \arctan(0)-\arctan(-1) \\ &= 0- \left( - \dfrac{\pi}{4} \right) \\ &= \dfrac{\pi}{4}. \blacksquare \end{array}$$

Section 5.8 #28: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right).$$ Solution: Since $\tanh(x)=\dfrac{\sinh(x)}{\cosh(x)}$, we may proceed with the quotient rule or we simply use the formula $\dfrac{\mathrm{d}}{\mathrm{d}x} \tanh(x) = \mathrm{sech}^2(x)$. We will do the latter: compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right) &= \dfrac{1}{\tanh(\frac{x}{2})} \dfrac{\mathrm{d}}{\mathrm{d}x} \tanh \left( \dfrac{x}{2} \right) \\ &= \dfrac{1}{\tanh(\frac{x}{2})} \mathrm{sech}^2 \left( \dfrac{x}{2} \right) \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{x}{2} \right] \\ &= \dfrac{1}{2} \dfrac{\coth(\frac{x}{2})}{\sinh(\frac{x}{2})} \dfrac{1}{\cosh^2(\frac{x}{2})} \\ &= \dfrac{1}{2\sinh(\frac{x}{2})\cosh(\frac{x}{2})}. \end{array}$$ If one notes that $2\sinh\left(\dfrac{x}{2} \right)\cosh \left( \dfrac{x}{2} \right) = \sinh(x)$, we may write $$\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right) = \dfrac{1}{\sinh(x)} = \mathrm{csch}(x). \blacksquare$$

Section 5.8 #30: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left[ x \cosh(x)-\sinh(x) \right].$$ Solution: Since $\cosh'(x)=\sinh(x)$ and $\sinh'(x)=\cosh(x)$, we use the product rule to compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} [ x \cosh(x) - \sinh(x) ] &= [\cosh(x) + x \sinh(x) ] - \cosh(x) \\ &= x \sinh(x). \blacksquare \end{array}$$

Section 5.8 #43: Calculate $$\displaystyle\int \cosh(2x) \mathrm{d}x.$$ Solution: Let $u=2x$ so that $\dfrac{1}{2} \mathrm{d}u=\mathrm{d}x$. Then, compute $$\begin{array}{ll} \displaystyle\int \cosh(2x) \mathrm{d}x &= \dfrac{1}{2} \displaystyle\int \cosh(u) \mathrm{d}u \\ &= \dfrac{1}{2} \sinh(u) + C \\ &= \dfrac{1}{2} \sinh(2x) + C. \blacksquare \end{array}$$

Section 5.8 #55: Calculate $$\displaystyle\int_0^{\ln(2)} \tanh(x) \mathrm{d}x.$$ Solution: Since $\tanh(x) = \dfrac{\sinh(x)}{\cosh(x)}$, let $u=\cosh(x)$ so that $\mathrm{d}u=\sinh(x)$. Also whenever $x=0$ we have $$u=\cosh(0)=\dfrac{e^0+e^{-0}}{2}=1,$$ and if $x=\ln(2)$ we have $$u=\cosh(\ln(2))=\dfrac{e^{\ln(2)}+e^{-\ln(2)}}{2}=\dfrac{2+\frac{1}{2}}{2}=\dfrac{5}{4}.$$ So compute $$\begin{array}{ll} \displaystyle\int_0^{\ln(2)} \tanh(x) \mathrm{d}x &= \displaystyle\int_0^{\ln(2)} \dfrac{\sinh(x)}{\cosh(x)} \mathrm{dx} \\ &= \displaystyle\int_1^{\frac{5}{4}} \dfrac{1}{u} \mathrm{d}u \\ &= \log(u) \Bigg|_1^{\frac{5}{4}} \\ &= \log \left( \dfrac{5}{4} \right) - \log(1) \\ &= \log\left( \dfrac{5}{4} \right). \blacksquare \end{array}$$

Section 5.8 #75: Calculate $$\displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x.$$ Solution: Note that $$\dfrac{1}{3-9x^2} = \dfrac{1}{(\sqrt{3})^2-(3x)^2}.$$ So let $u=3x$ so that $\dfrac{1}{3} \mathrm{d}u = \mathrm{d}x$. Since $$\displaystyle\int \dfrac{1}{a^2-x^2} \mathrm{d}x = \dfrac{1}{a} \arctan\left( \dfrac{x}{a} \right) + C,$$ we now calculate $$\begin{array}{ll} \displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x &= \displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x \\ &= \displaystyle\int \dfrac{1}{\sqrt{3}^2-(3x)^2} \mathrm{d}x \\ &= \dfrac{1}{3} \displaystyle\int \dfrac{1}{\sqrt{3}^2-u^2} \mathrm{d}u \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh} \left( \dfrac{u}{\sqrt{3}} \right) + C \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh}\left( \dfrac{3x}{\sqrt{3}} \right) + C \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh} \left( \sqrt{3}x \right) + C. \blacksquare \end{array}$$

Section 5.8 #86: Calculate $$\displaystyle\int_0^1 \dfrac{1}{\sqrt{25x^2+1}} \mathrm{d}x.$$ Solution: Note that $$\dfrac{1}{\sqrt{25x^2+1}} = \dfrac{1}{\sqrt{(5x)^2+1}}.$$ So letting $u=5x$ implies $\dfrac{1}{5} \mathrm{d}u = \mathrm{d}x$ and if $x=0$ then $u=0$ and if $x=1$ then $u=5$. Therefore compute $$\begin{array}{ll} \displaystyle\int_0^1 \dfrac{1}{\sqrt{25x^2+1}} \mathrm{d}x &= \displaystyle\int_0^1 \dfrac{1}{\sqrt{(5x)^2+1}} \mathrm{d}x \\ &= \dfrac{1}{5} \displaystyle\int_0^5 \dfrac{1}{\sqrt{u^2+1}} \mathrm{d}u \\ &= \dfrac{1}{5} \mathrm{arcsinh}(u) \Bigg|_0^5 \\ &= \dfrac{1}{5} \left[ \mathrm{arcsinh}(5) - \mathrm{arcsinh}(0) \right] \\ &= \dfrac{\mathrm{arcsinh}(5)}{5}. \blacksquare \\ \end{array}$$