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7.4  p. 582:   3,  5, 11, 13, 15, 19, 27, 29, 31, 39, 49.
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\SN{\jobname}

\np $\int^2_{-2}\Frac{dx}{4+x^2}=
\frac12\arctan(\frac x2)\Big\vert^2_{-2}=
\frac12\br2{\arctan 1-\arctan(-1)}=\frac12\br2{\frac\pi4-(-\frac\pi4}=
\frac\pi4$.

\np$\int^{3/2}_0\Frac{dx}{\sqrt{9-x^2}}=
\arcsin(\frac x3)\Big\vert^{3/2}_0=
\arcsin(\frac 12)-\arcsin 0=\frac\pi6-0=\frac\pi6
$.

\np$\int\Frac{\sqrt{y^2-49}}{y}\>dy$.
Substitute $y=7\sec u$, $dy=7\sec u\>\tan u\  du$, so
$\int\Frac{\sqrt{y^2-49}}{y}\>dy=
\int\Frac{49\tan u\sec u\tan u}{7\sec u}\>du=
7\int \tan^2u\>du=
7\int(\sec^2 u-1)du=
7\br1{\tan u - u}+C=
7\br2{\Frac{\sqrt{y^2-49}}{7}-\arc\sec(\frac y7)}+C
$.
Use the triangle\hskip4pt $\triangle{7}{\sqrt{y^2-49}}{y}$.

\np$\int\Frac{dx}{x^2\sqrt{x^2-1}}$.
Substitute $x=\sec u$, $du=\sec u\>\tan u\ du$, so
$\int\Frac{dx}{x^2\sqrt{x^2-1}}=
\int\Frac{\sec u\>\tan u}{\sec^2u\>\tan u}\> du=
\int\Frac 1{\sec u}\>du=\int\cos u\>du=
\sin u+C=\Frac{\sqrt{x^2-1}}{x}+C
$.
Use the triangle $\triangle{1}{\sqrt{x^2-1}}{x}$.

\np$\int\Frac{x^{3}\>dx}{\sqrt{x^2+4}}$.
Substitute $x=2\tan u$, $dx=2\sec^2u\> du$, so
$\int\Frac{x^{3}\>dx}{\sqrt{x^2+4}}=
\Frac{16\tan^3u\>\sec^2u}{2\sec u}\>du=
8\int\tan^3u\> \sec u\ du=
8\int\tan^2u(\tan u\>\sec u)\>du=
8\int(\sec^2 u-1)(\tan u\> \sec u)\>du$.
Now substitute $w=\sec u$, $dw=\tan u\>\sec u\ dw$,
so $8\int(\sec^2 u-1)(\tan u\> \sec u)\>du=
8\int (w^2-1)dw=8\br1{\frac {w^3}{3}-w}+C$.
Furthermore, $w=\sec u$: use the triangle
\hskip20pt
$\triangle{2}{x}{\sqrt{x^2+4}}$ \hskip40pt
so $w=\frac{\sqrt{x^2+4}}{2}$.
Hence
$\int\Frac{x^{3}\>dx}{\sqrt{x^2+4}}=
8\Frac{\br1{\sqrt{x^2+4}}^3}{3\cdot 2^3}-
8\Frac{\sqrt{x^2+4}}{2}+C=
\Frac{\br1{x^2+4}^{3/2}}{3}-4\sqrt{x^2+4}+C
$.

\np$\int^{\sqrt3/2}_0\Frac{4x^2dx}{(1-x^2)^{3/2}}$.
Substitute $x=\sin u$, $dx=\cos u\> du$, so
$\int^{\sqrt3/2}_0\Frac{4x^2dx}{(1-x^2)^{3/2}}=
\int^{\pi/3}_{0}\Frac{4\sin^2u\>\cos u\ du}{\cos^{3} u}=
4\int^{\pi/3}_{0}\tan^2 u\> du=4\int^{\pi/3}_{0}\br1{\sec^2 u -1}\>du=
4\br2{\tan u - u}\Big\vert^{\pi/3}_0=4\br1{\tan(\pi/3)-\frac\pi3}-
4\br1{\tan 0 - 0}=4\sqrt 3-\frac{4\pi}3$.


\np$\int\Frac{v^2 dv}{(1-v^2)^{5/2}}$. 
Substitute $v=\sin u$, $dv=\cos u\> du$, so
$\int\Frac{v^2 dv}{(1-v^2)^{5/2}}=
\int\Frac{\sin^2 u\>\cos u}{\cos^5 u}\>du=
\int\tan^2u\>\sec^2u\ du$.
Substitute $w=\tan u$, $dw=\sec^2u\>du$, so
$\int\Frac{v^2 dv}{(1-v^2)^{5/2}}=\int w^2\>dw=\Frac{w^3}3+C$.
Now $w=\tan u$ and using the triangle \hskip10pt
$\triangle{\sqrt{1-v^2}}{v}{1}$\hskip40pt
we see $w=\Frac{v}{\sqrt{1-v^2}}$, so
$\int\Frac{v^2 dv}{(1-v^2)^{5/2}}=
\frac13\br2{\Frac{v}{\sqrt{1-v^2}}}^3+C$.


\np$\int^{\ln 4}_0\Frac{e^t dt}{\sqrt{e^{2t}+9}}$.
Substitute $x=e^t$, $dx=e^t dt$, so
$\int^{\ln 4}_0\Frac{e^t dt}{\sqrt{e^{2t}+9}}=
\int^4_1\Frac{dx}{\sqrt{x^2+9}}$.
Now substitute $x=3\tan u$, $dx=3\sec^2u\> du$ so
$\int^{\ln 4}_0\Frac{e^t dt}{\sqrt{e^{2t}+9}}=
\int^{\arctan(4/3)}_{\arctan(1/3)}\Frac{3\sec^2u\> du}{3\sec u}=
3\int^{\arctan(4/3)}_{\arctan(1/3)}\sec u\> du=
\ln\vert \tan u+\sec u\vert \Big\vert^{\arctan(4/3)}_{\arctan(1/3)}$.
Use the triangle\hskip 30pt\triangle{3}{x}{\sqrt{x^2+9}}\hskip60pt
Filling in the numbers for $\arctan(43)$\quad
\triangle{3}{4}{5}\hskip40pt
so $\sec\br1{\arctan(4/3)}=\frac 53$
and $\tan\br1{\arctan(4/3)}=\frac43$.
The numbers for $\arctan(1/3)$\quad
\triangle{3}{1}{\sqrt{10}}\hskip40pt
so $\sec\br1{\arctan(1/3)}=\frac {\sqrt{10}}3$
and $\tan\br1{\arctan(1/3)}=\frac13$.
Hence
$\int^{\ln 4}_0\Frac{e^t dt}{\sqrt{e^{2t}+9}}=
\ln\br1{\frac 93}-\ln\br1{\frac{\sqrt{10}}3+\frac13}=
\ln 9-\ln(1+\sqrt{10})$.

\np$\int^{1/4}_{1/12}\Frac{2 dt}{\sqrt t+ 4t\sqrt t}$.
Let $u=2\sqrt t$, $du = \Frac{dt}{\sqrt t}$, so
$\int^{1/4}_{1/12}\int^{1}_{1/\sqrt3}\Frac{2 dt}{\sqrt t+ 4t\sqrt t}=
\int^{1}_{1/\sqrt3}\Frac{2du}{1+u^2}=
2\arctan u\Big\vert^{1}_{1/\sqrt3}=
2\br1{\arctan 1-\arctan(1/\sqrt3)}=
2\br1{\frac\pi4-\frac\pi6}=\Frac\pi6$.

\np$(x^2+4)y^\prime = 3$, $y(2)=0$.
First separate the differential equation:
$dy=3\Frac{dx}{x^2+4}$, so
$\int dy=3\int \Frac{dx}{x^2+4}$ or
$y=\frac32\arctan(\frac x2)+C$.
We already have $y$ as a function of $x$ so we can skip this step.
To evaluate $C$,
$0=y(2)=\frac 32\arctan(\frac 22)+C=\frac 32\arctan(1)+C=
\Frac{3\pi}{8}+C$ so $C=-\Frac{3\pi}8$ or
$y=\frac 32\arctan(\frac x2)-\frac{3\pi}8$.

\np$\int\Frac{dt}{\sin t -\cos t}$.
By the book $z=\tan(t/2)$, $dz=\frac12\sec^2(t/2)\> dt=
(1+\tan^2(t/2))dt$ or
$dt=\Frac{2dz}{1+z^2}$.
Moreover
$\cos t=\Frac{1-z^2}{1+z^2}$; $\sin t=\Frac{2z}{1+z^2}$.
Hence
$\int\Frac{dt}{\sin t -\cos t}=
2\int\Frac{\frac{dz}{1+z^2}}{\frac{2z}{1+z^2}-\frac{1-z^2}{1+z^2}}=
2\int\Frac{dz}{-1+2z+z^2}$.
Next we need partial fractions.
$z^2+2z-1$ is reducible.
It is not easy to see how to factor it, but the quadratic formula locates
the roots as $\Frac{-2\pm\sqrt{2^2-4(1)(-1)}}{2}=
\Frac{-2\pm\sqrt{4+4}}{2}=-1\pm\sqrt2$.
Hence
$z^2+2z-1=\br1{z-(-1+\sqrt2)}\br1{z-(-1-\sqrt2)}$.
Then
$\Frac{1}{z^2+2z-1}=\Frac A{z-(-1+\sqrt2)}+\Frac B{z-(-1-\sqrt2)}$,
or $1=A\br1{z-(-1-\sqrt2)}+B\br1{z-(-1+\sqrt2)}$.
Plug in $z=-1+\sqrt 2$; $1=A\br1{(-1+\sqrt 2)-(-1-\sqrt 2)}=
A\cdot2\sqrt2$, or $A=\Frac{1}{2\sqrt2}$.
Plug in $z=-1-\sqrt 2$; $1=B\br1{(-1-\sqrt 2)-(-1+\sqrt 2)}$, or
$B=\Frac{-1}{2\sqrt2}$.
Hence $\int\Frac{dz}{z^2+2z-1}=
\Frac{1}{2\sqrt2}\int\br2{
\Frac 1{z-(-1+\sqrt2)}-\Frac{1}{z-(-1-\sqrt2)}}\>dz=
\Frac{1}{2\sqrt2}\br2{
\ln\big\vert z-(-1+\sqrt2)\big\vert-
\ln\big\vert z-(-1-\sqrt2)\big\vert}+C=
\Frac1{2\sqrt2}\ln
\Bigg\vert\Frac{z-(-1+\sqrt2)}{z-(-1-\sqrt2)}\Bigg\vert+C
$.
Since $z=\tan(t/2)$,
$\int\Frac{dt}{\sin t -\cos t}=
\Frac1{\sqrt2}\ln
\Bigg\vert\Frac{\tan(t/2)+1-\sqrt2}
{\tan(t/2)+1+\sqrt2}\Bigg\vert+C$.


\tg

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