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Section 5.6 #56: Find the derivative of $25\arcsin\left(\dfrac{x}{5} \right)-x\sqrt{25-x^2}$.
Solution: Calculate $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} \left[25\arcsin\left(\dfrac{x}{5}\right)-x\sqrt{25-x^2} \right] &= \left[ 25 \dfrac{1}{\sqrt{1-(\frac{x}{5})^2}} \dfrac{1}{5} \right] - \left[ (1)\sqrt{25-x^2} + \dfrac{x}{\sqrt{25-x^2}} (-2x) \right] \\ &=\dfrac{25}{\sqrt{25-x^2}} - \sqrt{25-x^2} - \dfrac{2x^2}{\sqrt{25-x^2}} \\ \end{array}$$

Section 5.7 #68(a): Show that $\displaystyle\int_0^1 \dfrac{4}{1+x^2} \mathrm{d}x = \pi$.
Solution: Compute $$\begin{array}{ll} \displaystyle\int_0^1 \dfrac{4}{1+x^2} \mathrm{d}x &= 4 \arctan(x) \Bigg|_0^1 \\ &= 4 \left[ \arctan(1) - \arctan(0) \right] \\ &= 4 \left[ \dfrac{\pi}{4} - 0 \right] \\ &= \pi. \end{array}$$

Section 5.8 #102: Let $x>0$ and $b>0$. Show that $$\displaystyle\int_{-b}^be^{xt} \mathrm{d}t = \dfrac{2\sinh(bx)}{x}.$$ Solution: Note that $t$ is the variable being integrated and so the variable "$x$" in the integral is treated as a constant. This means we should use a $u$-substitution with $u=xt$ so that $\mathrm{d}u = x \mathrm{d}t$ (note we didn't do $\mathrm{d}u=t \mathrm{d}x$ which is not "untrue", but it is not useful for our purposes because the integral we're computing has a "$\mathrm{d}t$" in it) and so $\dfrac{1}{x} \mathrm{d}u = \mathrm{d}t$. Consequently, if $t=-b$ then $u=-xb$ and if $t=b$ then $u=xb$. Now compute $$\begin{array}{ll} \displaystyle\int_{-b}^b e^{xt} \mathrm{d}t &= \dfrac{1}{x} \displaystyle\int_{-xb}^{xb} e^u \mathrm{d}u \\ &= \dfrac{1}{x} e^u \Bigg|_{-xb}^{xb} \\ &= \dfrac{e^{xb} - e^{-xb}}{x}. \end{array}$$ But recall the definition of $\sinh$ which was $\sinh(z)=\dfrac{e^z-e^{-z}}{2}$. From this we see that $2\sinh(z)=e^z-e^{-z}$. Therefore we see, using $z=xb$, that $$\displaystyle\int_{-b}^b e^{xt}\mathrm{d}t = \dfrac{e^{xb}-e^{-xb}}{x} = \dfrac{2\sinh(xb)}{x},$$