\input homework_macros.tex

%\distiller
\printer

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

\pn3.$\int3\sqrt{\sin\> v\ }\cos\> v\ dv$: substitute
$w=\sin\>v$, $dw=\cos\>v\ dv$ so
$\int 3\sqrt w\ dw=3\int w^{1/2}\ dw=3\frac{w^{1+1/2}}{1+1/2}+C=
2\ w^{\frac 32}+C=2 (\sin\>v)^{\frac32}+C$.

\pn7.$\int\frac{dx}{\sqrt x(\sqrt x+1)}$: substitute 
$u=\sqrt x +1$, $du= d(x^{1/2})=\frac 12x^{\frac12-1}dx=
\frac 1{2\sqrt x}dx$ so
$\int\frac{dx}{\sqrt x(\sqrt x+1)}=\int\frac2u\ du=2\ln\vert u\vert +C=
2\ln(\sqrt x+1)+C$.
The absolute value signs would be correct also, 
but they can be dropped since $\sqrt x+1>0$

\pn11.$\int e^\theta\csc(e^\theta+1)\>d\theta$: substitute
$u=e^\theta+1$, $du=e^\theta\>d\theta$.
$\int e^\theta\csc(e^\theta+1)\>d\theta=\int\csc u\ du=
-\ln\bigl\vert\csc\>u+\cot\>u\bigr\vert+C=
-\ln\bigl\vert\csc\br0{e^\theta+1}+\cot\br0{e^\theta+1}\bigr\vert+C$.

\pn17.$\int_0^{\sqrt{\ln2}} 2xe^{x^2}\>dx$: substitute $u=x^2$,
$du=2x\>dx$ so
$\int_0^{\sqrt{\ln2}} 2xe^{x^2}\>dx=\int_0^{\ln2} e^u\>du=
e^u\Big\vert^{\ln2}_0=e^{\ln2}-e^0=2-1=1$.

\pn31.$\int\frac{6dx}{x\sqrt{25x^2-1}}$: substitute
$u=5x$, $du=5dx$ or $\frac65du=6dx$ so
$\int\frac{6dx}{x\sqrt{25x^2-1}}=
\int\frac{\frac65du}{\frac u5\sqrt{u^2-1}}=
6\int\frac{du}{u\sqrt{u^2-1}}=
6\arc\sec\vert u\vert+C=6\arc\sec\vert 5x\vert+C$.

\pn35.$\int_1^{e^{\pi/3}}\frac{dx}{x\cos(\ln\>x)}$: substitute
$u=\ln x$, $du=\frac{dx}{x}$.
$\int_1^{e^{\pi/3}}\frac{dx}{x\cos(\ln\>x)}=
\int_0^{\pi/3}\frac{du}{\cos(u)}=
\int_0^{\pi/3}\sec\>u\ du=
\ln\vert\sec\>u+\tan\>u\vert\ \Big\vert^{\pi/3}_0=
\ln\br1{\sec(\pi/3)+\tan(\pi/3)}-
\ln\br1{\sec(0)+\tan(0)}=
\ln\br1{2+\sqrt3}-
\ln\br1{1+0}=
\ln(2+\sqrt3)
$

\pn41.$\int\frac{dx}{(x+1)\sqrt{x^2+2x}}$.
Completing the square: $x^2+2x=(x+1)^2-1$.
Substitute $u=x+1$, $du=dx$, so
$\int\frac{dx}{(x+1)\sqrt{x^2+2x}}=
\int\frac{du}{u\sqrt{u^2-1}}=\arc\sec\vert u\vert+C
=\arc\sec\vert x+1\vert+C$.

\pn47.$\int\frac{x\ dx}{x+1}$.
Polynomial long division
$\frac x{x+1}=1-\frac 1{x+1}$ so
$\int\frac{x\ dx}{x+1}=
\int\br1{1-\frac 1{x+1}}\ dx=
x-\ln\vert x+1\vert+C$.

\pn49.$\int\frac{2x^3}{x^2-1}\>dx$.
Polynomial long division yields
$\frac{2x^3}{x^2-1}=2x+\frac{2x}{x^2-1}$ so
$\int\frac{2x^3}{x^2-1}\>dx=\int2x\>dx+\int\frac{2x}{x^2-1}\>dx=
x^2+\ln\vert x^2-1\vert+C$.

\pn53.$\int\frac{1-x}{\sqrt{1-x^2}}\>dx=
\int\frac{1}{\sqrt{1-x^2}}\>dx-\int\frac{x}{\sqrt{1-x^2}}\>dx$.
Do the integral $\int\frac{x}{\sqrt{1-x^2}}\>dx$ by
substitution $u=1-x^2$, $du=-2x\>dx$, so
$\int\frac{x}{\sqrt{1-x^2}}\>dx=-\frac12\int\frac{du}{\sqrt u}=
-\frac12\int u^{-1/2}du=-\frac12 \frac{u^{1-1/2}}{1-1/2}+C=
-\frac12 \frac{u^{1/2}}{1/2}+C=-\sqrt u+C=-\sqrt{1-x^2}+C$.
Hence 
$\int\frac{1}{\sqrt{1-x^2}}\>dx-\int\frac{x}{\sqrt{1-x^2}}\>dx=
\arcsin x-(-\sqrt{1-x^2})+C=
\arcsin x+\sqrt{1-x^2}+C$.

\pn57.$\int\frac 1{1+\sin x}dx=
\int\frac 1{1+\sin x}\frac{1-\sin x}{1-\sin x}dx=
\int\frac{1-\sin x}{1-\sin^2x}dx=
\int\frac{1-\sin x}{\cos^2 x}dx=\int\sec^2x\>dx-\int\sec x\tan x\>dx=
\tan x-\sec x+C$.

\pn65.$\int_{\pi/2}^{\pi}\sqrt{1+\cos 2t\>}\>dt$.
The Double Angle Formula yields
$2\cos^2t=1+\cos 2t$, so
$\int_{\pi/2}^{\pi}\sqrt{1+\cos 2t\>}\>dt=
\int_{\pi/2}^{\pi}\sqrt{2\cos^2t\>}\>dt=
\sqrt2\int_{\pi/2}^{\pi}\vert\cos t\vert\>dt=
-\sqrt2\int_{\pi/2}^{\pi}\cos t\>dt=
-\sqrt2\sin t\Big\vert^{\pi}_{\pi/2}=-\sqrt2\br1{\sin \pi - \sin(\pi/2)}=
-\sqrt2(0-1)=\sqrt{2}$

\end