\input homework_macros.tex

%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}
\def\arc#1{{\rm arc{#1}}\>}

\SN{6.9}
\pn1. $\dd{\arccos{x^2}}x=\frac{\dd{x^2}x}{\sqrt{1-(x^2)^2}}=
\frac{2x}{\sqrt{1-x^4}}$.

\pn3. $\dd{\arcsin{\sqrt2\>t}}t=
\frac{1}{\sqrt{1-(\sqrt2\>t)^2}}\dd{\sqrt2\>t}t=
\frac{\sqrt2}{\sqrt{1-2t^2}}$.

\pn5. $\dd{\arc{sec}{(2s+1)}}s=
\frac{1}{\vert 2s+1\vert\sqrt{(2s+1)^2-1}}\dd{2s+1}s=
\frac{2}{\vert 2s+1\vert\sqrt{4s^2+4s}}=
\frac{1}{\vert 2s+1\vert\sqrt{s^2+s}}$.

\pn8. $\dd{\arc{csc}(\frac x2)}x=
-\frac{1}{\vert \frac x2\vert\sqrt{(\frac x2)^2-1}}\dd{\frac x2}{x}=
-\frac{\frac 12}{\vert \frac x2\vert\sqrt{\frac {x^2}4-1}}=
\frac{-8}{\vert x\vert\sqrt{x^2-4}}
$.

\pn17.
$\dd{s\sqrt{1-s^2}+\arccos s}s=
1\cdot\sqrt{1-s^2}+s(-2s)\frac 12(1-s^2)^{-1/2}-\frac 1{\sqrt{1-s^2}}=$
\par$
\frac{1-s^2}{\sqrt{1-s^2}}+\frac{-s^2}{\sqrt{1-s^2}}-\frac1{\sqrt{1-s^2}}=
\frac{-2s^2}{\sqrt{1-s^2}}
$.

\pn23.
$\int \frac{dx}{\sqrt{9-x^2}}$.
Substitute $3u=x$; $3du=dx$, so
$\int \frac{dx}{\sqrt{9-x^2}}=\int\frac{3du}{\sqrt{9-9u^2}}=
\int \frac{du}{\sqrt{1-u^2}}=\arcsin u+C=
\arcsin(\frac x3)+C$.

\pn25.
$\int\frac{dx}{17+x^2}$.
Substitute $\sqrt{17}\> u=x$ so
$\sqrt{17}\> du=dx$ and
$\int\frac{dx}{17+x^2}=\int\frac{\sqrt{17}\> du}{17+17u^2}=
\int\frac{\sqrt{du}}{1+1u^2}=\arctan u+C=
\arctan(\frac{x}{\sqrt{17}})+C$.

\pn27.
$\int\frac{dx}{x\sqrt{25x^2-2}}$.
Substitute $\sqrt2\>u=5x$; $\sqrt2\>du=5dx$.
Hence 
$\int\frac{dx}{x\sqrt{25x^2-2}}=$\par$
\int\frac{\frac{\sqrt2\> du}5}{
(\frac{\sqrt2\>u}{5})\sqrt{2u^2-2}}=
\frac1{\sqrt2}\int\frac{du}{u\sqrt{u^2-1}}=
\cases{\frac1{\sqrt2}\arc{sec} u+C&if $u>1$\cr
\frac{-1}{\sqrt2}\arc{sec} u+C&if $u<1$\cr}=$\par$
\frac1{\sqrt2}\arc{sec} \vert u\vert+C=
\frac1{\sqrt2}\arc{sec}\bigl(\big\vert\frac{\sqrt2\>x}{5}\big\vert\bigr)+C$.

\pn29.
$\int_0^1\frac{4ds}{\sqrt{4-s^2}}$. 
Substitute $2u=s$, ($2du=ds$) so
$\int_0^1\frac{4ds}{\sqrt{4-s^2}}=
\int_0^{\frac12}\frac{8du}{\sqrt{4-4u^2}}=$\par$
4\int_0^{\frac12}\frac{du}{\sqrt{1-u^2}}=
4\arcsin(u)\Big\vert_0^{1/2}=4(\frac\pi6-0)=\frac{2\pi}3$.

\pn31.
$\int_0^2\frac{dt}{8+2t^2}$.  
Substitute $2u=t$ ($2du=dt$) so
$\int_0^2\frac{dt}{8+2t^2}=
\int_0^1\frac{2du}{8+8t^2}=
\frac14\int_0^1\frac{du}{1+t^2}=
\frac{\arctan u}{4}\Big\vert_0^1=\frac{\arctan 1-\arctan 0}{4}=
\frac{\frac\pi4-0}4=\frac\pi{16}$.

\pn33.
$\int_{-1}^{-\frac{\sqrt2}2}\frac{dy}{y\sqrt{4y^2-1}}$.
Substitute $u=2y$ ($du=2dy$).
$\int_{-1}^{-\frac{\sqrt2}2}\frac{dy}{y\sqrt{4y^2-1}}=
\int_{-2}^{-\sqrt2}\frac{\frac{du}2}{\frac u2\sqrt{u^2-1}}$\par$
\int_{-2}^{-\sqrt2}\frac{du}{u\sqrt{u^2-1}}=
\arc{sec}\vert u\vert\Big\vert_{-2}^{-\sqrt2}=
\arc{sec}\sqrt2-\arc{sec}2=\frac\pi4-\frac\pi3=-\frac{\pi}{12}
$.

\pn35.$\int\frac{3dr}{\sqrt{1-4(r-1)^2}}$. Substitute $u=2(r-1)$
($du=2dr$).
$\int\frac{3dr}{\sqrt{1-4(r-1)^2}}=
\int\frac{\frac 32du}{\sqrt{1u^2}}=\frac32\arcsin u+C=
\frac 32\arcsin\br2{2(r-1)}+C$.

\pn37.
$\int\frac{dx}{2+(x-1)^2}$.
Substitute $\sqrt2\>u=(x-1)$ ($\sqrt2\>du=dx$.
$\int\frac{dx}{2+(x-1)^2}=
\int\frac{\sqrt2\>du}{2+2u^2}=\frac{\sqrt2}2\int\frac{du}{1+u^2}=
\frac{\sqrt2}2\arctan u+C=
\frac{\sqrt2}2\arctan\br2{\frac{x-1}{\sqrt2}}+C$.

\pn39.
$\int\frac{dx}{(2x-1)\sqrt{(2x-1)^2-4}}$.
Substitute $2u=2x-1$ ($2du=2dx$ or $du=dx$).
$\int\frac{dx}{(2x-1)\sqrt{(2x-1)^2-4}}=
\int\frac{du}{(2u)\sqrt{(2u)^2-4}}=
\frac{1}{2\cdot2}\int\frac{du}{u\sqrt{(u)^2-1}}=
\frac14\int\arc{sec}\vert u\vert+C=
\frac14\int\arc{sec}\Bigl\vert{\frac{2x-1}2}\Bigr\vert+C$.

\pn41.
$\int_{-\frac\pi2}^{\frac\pi2}\frac{2\cos\theta\>d\theta}
{1+(\sin\theta)^2}$.
Substitute $u=\sin\theta$ ($du=\cos\theta\>d\theta$).
$\int_{-\frac\pi2}^{\frac\pi2}\frac{2\cos\theta\>d\theta}
{1+(\sin\theta)^2}=
\int_{-1}^{1}\frac{2du}{1+u^2}=2\arctan u\Big\vert^1_{-1}=
2\br2{\frac\pi4-(-\frac\pi4)}=2\cdot\frac\pi2=\pi$.

\pn43.
$\int_0^{\ln\sqrt 3}\frac{e^xdx}{1+e^{2x}}$.
Substitute $u=e^x$ ($du=e^xdx$).
$\int_0^{\ln\sqrt 3}\frac{e^xdx}{1+e^{2x}}=
\int_1^{\sqrt3}\frac{du}{1+u^2}=
\arctan u\Big\vert^{sqrt3}_1=\arctan\sqrt3-\arctan1=
\frac\pi3-\frac\pi4=\frac\pi{12}$.

\pn45.
$\int\frac{ydy}{\sqrt{1-y^4}}$.
Substitute $u=y^2$ ($du=2ydy$).
$\int\frac{ydy}{\sqrt{1-y^2}}=\int\frac{\frac{du}2}{\sqrt{1-u^2}}=
=\frac12\arcsin u+C=\frac12\arcsin(y^2)+C$.

\pn47.
$\int\frac{dx}{\sqrt{-x^2+4x-3}}$.
Complete the square:
$-x^2+4x-3=-(x-2)^2+1$, so substitute $u=x-2$ ($du=dx$).
$\int\frac{dx}{\sqrt{-x^2+4x-3}}=
\int\frac{du}{\sqrt{1-u^2}}=\arcsin u+C=
\arcsin(x-2)+C$.

\pn49.
$\int_{-1}^0\frac{6dt}{\sqrt{3-2t-t^2}}$.
Complete the square: $3-2t-t^2=-(t+1)^2+4$ so substitute
$2u=t+1$ ($2du=dt$).
$\int_{-1}^0\frac{6dt}{\sqrt{3-2t-t^2}}=
\int_{0}^{\frac12}\frac{12du}{\sqrt{4-4u^2}}=
\frac{12}2\int_{0}^{\frac12}\frac{du}{\sqrt{1-u^2}}=
6\arcsin u\Big\vert^{\frac12}_0=6(\frac\pi6-0)=\pi$.

\pn51.
$\int\frac{dy}{y^2-2y+5}$.
Complete the square: $y^2-2y+5=(y-1)^2+4$ so substitute
$2u=y-1$ ($2du=dy$).
$\int\frac{dy}{y^2-2y+5}=
\int\frac{2du}{4u^2+4}=\frac24\int\frac{du}{1+u^2}=
\frac12\arctan u+C=
\frac12\arctan\br2{\frac{y-1}2}+C$.

\pn53.
$\int_1^2\frac{8dx}{x^2-2x+2}$.
Complete the square: $x^2-2x+2=(x-1)^2+1$ so substitute
$u=x-1$ ($du=dx$).
$\int_1^2\frac{8dx}{x^2-2x+2}=
\int_0^1\frac{8du}{u^2+1}=8\arctan u\Big\vert_0^1=
8(\arctan 1-\arctan 0)=8(\frac\pi4-0)=2\pi$.

\pn55.
$\int\frac{dx}{(x+1)\sqrt{x^2+2x}}$.
Complete the square: $x^2+2x=(x+1)^2-1$ so substitute
$u=x+1$ ($du=dx$).
$\int\frac{dx}{(x+1)\sqrt{x^2+2x}}=
\int\frac{du}{u\sqrt{u^2-1}}=\arc{sec}\vert u\vert+C=
\arc{sec}\vert x+1\vert+C$.

\pn57.
$\int \frac{e^{\arcsin x}dx}{\sqrt{1-x^2}}$.
Substitute $u=\arcsin x$ ($du=\frac{dx}{\sqrt{1-x^2}}$).
$\int \frac{e^{\arcsin x}dx}{\sqrt{1-x^2}}=
\int {e^udu}=e^u+C=e^{\arcsin x}+C$.

\pn59.
$\int \frac{(\arcsin x)^2dx}{\sqrt{1-x^2}}$.
Substitute $u=\arcsin x$ ($du=\frac{dx}{\sqrt{1-x^2}}$).
$\int \frac{(\arcsin x)^2dx}{\sqrt{1-x^2}}=
\int u^2du=\frac{u^3}3+C=
\frac{(\arcsin x)^3}3+C$.

\pn61.
$\int\frac{dy}{(\arctan y)(1+y^2)}$.
Substitute $u=\arctan y$ ($du=\frac{dy}{1+y^2}$).
$\int\frac{dy}{(\arctan y)(1+y^2)}=
\int \frac {du}u=\ln\vert u\vert+C=
\ln\vert\arctan x\vert+C$.

\pn63.
$\int_{\sqrt2}^{2}\frac{\sec^2(\arc{sec} x)dx}{x\sqrt{x^2-1}}$.
First simplify: $\sec(\arc{sec} x)=x$ so 
$\int_{\sqrt2}^{2}\frac{\sec^2(\arc{sec} x)dx}{x\sqrt{x^2-1}}=
\int_{\sqrt2}^{2}\frac{x^2dx}{x\sqrt{x^2-1}}=
\int_{\sqrt2}^{2}\frac{xdx}{\sqrt{x^2-1}}$.
Now substitute $u=x^2-1$ ($du=2x\>dx$).
$\int_{\sqrt2}^{2}\frac{\sec^2(\arc{sec} x)dx}{x\sqrt{x^2-1}}=
\int_{1}^{3}\frac12\frac{du}{\sqrt u}=
\int_{1}^{3}\frac12 u^{-1/2}du=\frac12\frac{u^{-1/2+1}}{-1/2+1}
\Big\vert^3_1=\frac12\frac{u^{1/2}}{1/2}\Big\vert^3_1=
u^{1/2}\Big\vert^3_1=\sqrt3-\sqrt1=\sqrt3-1$.


\pn65.
$\displaystyle\lim_{x\to0}\frac{\arcsin 5x}x$.
Since $\arcsin(5\cdot0)=\arcsin(0)=0$ this limit has the form
$\frac00$.
By l'H\^opital's Rule,
$\displaystyle\lim_{x\to0}\frac{\arcsin 5x}x=
\displaystyle\lim_{x\to0}\frac{\dd{\arcsin 5x}x}1=
\displaystyle\lim_{x\to0}\frac{\frac{\dd{5x}x}{\sqrt{1-(5x)^2}}}1=
\displaystyle\lim_{x\to0}\frac{5}{\sqrt{1-25x^2}}=
\frac{5}{\sqrt{1-25\cdot0^2}}=5$.


\pn73. Solve $\dd yx=\frac1{\sqrt{1-x^2}}$, $y(0)=0$.
First solve the differential equation:
$y=\arcsin x+C$.
Then solve $0=y(0)=\arcsin(0) +C$.
Since $\arcsin(0)=0$, $C=0$ and the solution is
$y=\arcsin x$.

\end

6.9  p. 518:   1,  3,  5,  8, 17, 23 - 65 (odd), 73
