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Section 5.6 #39: Compute $\dfrac{\mathrm{d}}{\mathrm{d}x} 2\arcsin(x-1)$.
Solution: Using the formula $\dfrac{\mathrm{d}}{\mathrm{d}x} \arcsin(x)=\dfrac{1}{\sqrt{1-x^2}}$ and the chain rule, we get
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} [2\arcsin(x-1)] &= 2 \left(\dfrac{1}{\sqrt{1-(x-1)^2}} \right) \dfrac{\mathrm{d}}{\mathrm{d}x} [x-1] \\
&= \dfrac{2}{\sqrt{1-(x-1)^2}} \\
&= \dfrac{2}{\sqrt{1-(x^2-2x+1)}} \\
&= \dfrac{2}{\sqrt{2x-x^2}} \\
&= \dfrac{2}{\sqrt{x(2-x)}}.
\end{array}$$
Note: any of the last 4 lines above would be considered a "correct" answer worth full credit on the exam
Section 5.6 #43: Compute $\dfrac{\mathrm{d}}{\mathrm{d}x} \arctan(e^x)$.
Solution: Using the formula $\dfrac{\mathrm{d}}{\mathrm{d}x} \arctan(x)=\dfrac{1}{x^2+1}$ and the chain rule, we get
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \arctan(e^x) &= \dfrac{1}{1+(e^x)^2} \dfrac{\mathrm{d}}{\mathrm{d}x} [e^x] \\
&= \dfrac{e^x}{1+e^{2x}}.
\end{array}$$
Section 5.7 #23: Compute $\displaystyle\int_0^{\frac{\sqrt{3}}{2}} \dfrac{1}{1+4x^2} \mathrm{d}x$.
Solution: This requires remembering the integral formula $\displaystyle\int \dfrac{1}{a^2+x^2} \mathrm{d}x = \dfrac{1}{a} \arctan \left( \dfrac{x}{a} \right) +C$ in the special case when $a=1$. First notice that the integrand can be rewritten:
$$\dfrac{1}{1+4x^2} = \dfrac{1}{1+(2x)^2},$$
which indicates we should use a $u$-substitution. Let $u=2x$ so that $\mathrm{d}u=2\mathrm{d}x$, or equivalently, $\dfrac{1}{2} \mathrm{d}u = \mathrm{d}x$. This causes the lower limit of integration $x=0$ to become $u=2(0)=0$ and the upper limit of integration $x=\dfrac{\sqrt{3}}{2}$ to become $u=2 \left( \dfrac{\sqrt{3}}{2} \right) = \sqrt{3}$. Now compute
$$\begin{array}{ll}
\displaystyle\int_0^{\frac{\sqrt{3}}{2}} \dfrac{1}{1+4x^2} \mathrm{d}x &= \displaystyle\int_0^{\frac{\sqrt{3}}{2}} \dfrac{1}{1+(2x)^2} \mathrm{d}x \\
&=\dfrac{1}{2} \displaystyle\int_0^{\sqrt{3}} \dfrac{1}{1+u^2} \mathrm{d}u \\
&= \dfrac{\arctan(u)}{2} \Bigg|_0^{\sqrt{3}} \\
&= \dfrac{1}{2} \arctan(\sqrt{3}) - \dfrac{\arctan(0)}{2} \\
&= \dfrac{1}{2} \left( \dfrac{\pi}{3} \right) - 0 \\
&= \dfrac{\pi}{6}.
\end{array}$$
Section 5.8 #30: Compute the derivative of $x\cosh(x)-\sinh(x)$.
Solution: Use the fact that $\dfrac{\mathrm{d}}{\mathrm{d}x} \cosh(x)=\sinh(x)$ and $\dfrac{\mathrm{d}}{\mathrm{d}x} \sinh(x)=\cosh(x)$ and the product rule to compute
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} [x\cosh(x)-\sinh(x)] &= [(1)\cosh(x) + x \sinh(x)] -\cosh(x) \\
&=\cosh(x)+x\sinh(x)-\cosh(x) \\
&= x\sinh(x).
\end{array}$$
Section 5.8 #49: Calculate $\displaystyle\int \dfrac{\cosh(x)}{\sinh(x)} \mathrm{d}x$.
Solution: Let $u=\sinh(x)$ so $\mathrm{d}u=\cosh(x) \mathrm{d}x$. Now use this substitution to compute
$$\begin{array}{ll}
\displaystyle\int \dfrac{\cosh(x)}{\sinh(x)} \mathrm{d}x &= \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= \ln(|u|) + C \\
&= \ln(|\sinh(x)|) +C.
\end{array}$$
Section 5.8 #70: Calculate the derivative of $\mathrm{arctanh}(\sin(2x))$.
Solution: Using the formula $\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{arctanh(x)} = \dfrac{1}{1-x^2}$ and the chain rule, compute
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{arctanh(\sin(2x))} &= \dfrac{1}{1-(\sin(2x))^2} \dfrac{\mathrm{d}}{\mathrm{d}x} [\sin(2x)] \\
&= \dfrac{2\cos(2x)}{1-\sin^2(2x)} \\
&= \dfrac{2\cos(2x)}{\cos^2(2x)} \\
&= \dfrac{2}{\cos(2x)} \\
&= 2\sec(2x).
\end{array}$$
Note: any of the last 4 lines above would constitute "full credit" for this problem on an exam.
Section 5.8 #86: Compute $\displaystyle\int_0^1 \dfrac{1}{\sqrt{25x^2+1}} \mathrm{d}x$.
Solution: First note that
$$\dfrac{1}{\sqrt{25x^2+1}} =\dfrac{1}{\sqrt{(5x)^2+1}},$$
so it is natural to use a $u$-substitution with $u=5x$ so that $\dfrac{1}{5} \mathrm{d}u=\mathrm{d}x$; the lower limit of integration $x=0$ becomes $u=5(0)=0$ and the upper limit of integration $x=1$ becomes $u=5(1)=5$. Use that $u$-substituion along with the integral formula $\displaystyle\int \dfrac{1}{\sqrt{x^2+a^2}} \mathrm{d}x=\mathrm{arcsinh}\left(\dfrac{x}{a}\right) + C$ to see
$$\begin{array}{ll}
\displaystyle\int_0^1 \dfrac{1}{\sqrt{25x^2+1}} \mathrm{d}x &= \displaystyle\int_0^1 \dfrac{1}{\sqrt((5x)^2+1)} \mathrm{d}x \\
&= \dfrac{1}{5} \displaystyle\int_0^5 \dfrac{1}{\sqrt{u^2+1}} \mathrm{d}u \\
&= \dfrac{1}{5} \mathrm{arcsinh}(u)\Bigg|_0^5 \\
&= \dfrac{1}{5} \left[ \mathrm{arcsinh}(5)-\mathrm{arcsinh}(0) \right] \\
&= \dfrac{1}{5} \mathrm{arcsinh}(5)
\end{array}$$
