$$
\begin{eqnarray}
\int \frac{1}{\cos{\left(\theta\right)}} \mathrm{d}\theta
&=&\int \sec{\left(\theta\right)} \mathrm{d}\theta
\;\cdots\;\sec{\left(\theta\right)}=\frac{1}{\cos{\left(\theta\right)}}
\\&=&\int \frac{1}{\cos{\left(\theta\right)}} \frac{\cos{\left(\theta\right)}}{\cos{\left(\theta\right)}} \mathrm{d}\theta
\\&=& \int \frac{\cos{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} \mathrm{d}\theta
\\&=& \int \frac{\cos{\left(\theta\right)}}{1-\sin^2{\left(\theta\right)}} \mathrm{d}\theta
\;\cdots\;\cos^2{\left(\theta\right)}=1-\sin^2{\left(\theta\right)}
\\&=& \int \frac{1}{2} \left\{
\frac{\cos{\left(\theta\right)}}{1+\sin{\left(\theta\right)}}
+\frac{\cos{\left(\theta\right)}}{1-\sin{\left(\theta\right)}}
\right\} \mathrm{d}\theta
\\&&\;\cdots\;\frac{\cos{\left(\theta\right)}}{1-\sin^2{\left(\theta\right)}}
=\frac{\alpha}{1+\sin{\left(\theta\right)}}
+\frac{\beta}{1-\sin{\left(\theta\right)}}
\\&&\;\cdots\;\cos{\left(\theta\right)}
=\alpha\left(1-\sin{\left(\theta\right)}\right)+\beta\left(1+\sin{\left(\theta\right)}\right)
=\left(\alpha+\beta\right)+\left(\alpha-\beta\right)\sin{\left(\theta\right)}
\\&&\;\cdots\;
\left\{
\begin{array}{l}
\cos{\left(\theta\right)}=\alpha+\beta
\\0=\left(\alpha-\beta\right)\sin{\left(\theta\right)}
\end{array}
\right.
\\&&\;\cdots\;\sin{\left(\theta\right)}は\thetaによっては0とはならないので,\alpha-\betaが常に0,よって\alpha=\beta.
\\&&\;\cdots\;従って\alpha+\beta=\alpha+\alpha=2\alpha=\cos{\left(\theta\right)}.
\\&&\;\cdots\;\alpha=\beta=\frac{1}{2}\cos{\left(\theta\right)}
\\&&\;\cdots\;\frac{\cos{\left(\theta\right)}}{1-\sin^2{\left(\theta\right)}}
=\frac{\frac{1}{2}\cos{\left(\theta\right)}}{1+\sin{\left(\theta\right)}}
+\frac{\frac{1}{2}\cos{\left(\theta\right)}}{1-\sin{\left(\theta\right)}}
\\&=& \frac{1}{2} \int \left\{
\frac{\cos{\left(\theta\right)}}{1+\sin{\left(\theta\right)}}
+\frac{\cos{\left(\theta\right)}}{1-\sin{\left(\theta\right)}}
\right\} \mathrm{d}\theta
\\&=& \frac{1}{2} \left\{
\int \frac{\cos{\left(\theta\right)}}{1+\sin{\left(\theta\right)}}\mathrm{d}\theta
+\int \frac{\cos{\left(\theta\right)}}{1-\sin{\left(\theta\right)}}\mathrm{d}\theta
\right\}
\\&=& \frac{1}{2} \left\{
\int \frac{\cos{\left(\theta\right)}}{u}\frac{1}{\cos{\left(\theta\right)}}\mathrm{d}u
+\int \frac{\cos{\left(\theta\right)}}{v}\frac{-1}{\cos{\left(\theta\right)}}\mathrm{d}v
\right\}
\\&&\;\cdots\;u=1+\sin{\left(\theta\right)},\frac{\mathrm{d}u}{\mathrm{d}\theta}=\cos{\left(\theta\right)},\mathrm{d}\theta=\frac{1}{\cos{\left(\theta\right)}}\mathrm{d}u
\\&&\;\cdots\;v=1-\sin{\left(\theta\right)},\frac{\mathrm{d}v}{\mathrm{d}\theta}=-\cos{\left(\theta\right)},\mathrm{d}\theta=\frac{-1}{\cos{\left(\theta\right)}}\mathrm{d}v
\\&=& \frac{1}{2} \left(
\int \frac{1}{u} \mathrm{d}u - \int \frac{1}{v} \mathrm{d}v
\right)
\\&=& \frac{1}{2} \left( \ln{\left|u\right|} - \ln{\left|v\right|} \right)\;\cdots\;\int \frac{1}{x} \mathrm{d}x=\ln{\left|x\right|}+C\;(C:積分定数)
\\&=& \frac{1}{2} \ln{\frac{\left|u\right|}{\left|v\right|}}\;\cdots\;\log{A}-\log{B}=\log{\frac{A}{B}}
\\&=& \frac{1}{2} \ln{\left|\frac{u}{v}\right|}\;\cdots\;\frac{\left|A\right|}{\left|B\right|}=\left|\frac{A}{B}\right|
\\&=& \frac{1}{2} \ln{\left|\frac{1+\sin{\left(\theta\right)}}{1-\sin{\left(\theta\right)}}\right|}
+ C \;\cdots\;C:積分定数
\\もしくは，
\\&=& \frac{1}{2} \ln{\left|
\frac{1+\sin{\left(\theta\right)}}{1-\sin{\left(\theta\right)}}
\right|}+ C
\;\cdots\;A\log{B}=\log{B^A}
\\&=& \frac{1}{2} \ln{\left|
\left\{
\frac{1}{\cos{\left(\theta\right)}}
+\tan{\left(\theta\right)}
\right\}^2
\right|}+C
\;\cdots\;\frac{
1+\sin{\left(\theta\right)}
}{
1-\sin{\left(\theta\right)}
}=\left\{
\frac{1}{\cos{\left(\theta\right)}}
+\tan{\left(\theta\right)}
\right\}^2
\;式変形は最後に記載
\\&=& \frac{1}{2} 2 \ln{\left|
\frac{1}{\cos{\left(\theta\right)}}
+\tan{\left(\theta\right)}
\right|}+C
\;\cdots\;\left(A^B\right)^\frac{1}{B}=A^\left(B\frac{1}{B}\right)=A^1=A
\\&=& \ln{\left|
\frac{1}{\cos{\left(\theta\right)}}
+\tan{\left(\theta\right)}
\right|}+C
\\&=& \ln{\left|
\sec{\left(\theta\right)}
+\tan{\left(\theta\right)}
\right|}+C
\;\cdots\;\frac{1}{\cos{\left(\theta\right)}}=\sec{\left(\theta\right)}
\end{eqnarray}
$$
$$
\begin{eqnarray}
\frac{
1+\sin{\left(\theta\right)}
}{
1-\sin{\left(\theta\right)}
}
&=&
\frac{1+\sin{\left(\theta\right)}}{1-\sin{\left(\theta\right)}}
\;\frac{1+\sin{\left(\theta\right)}}{1+\sin{\left(\theta\right)}}
\\&=&
\frac{
\left\{1+\sin{\left(\theta\right)}\right\}^2
}{
1-\sin^2{\left(\theta\right)}
}
\;\cdots\;(A+B)(A-B)=A^2-B^2
\\&=&
\frac{
1+2\sin{\left(\theta\right)}+\sin^2{\left(\theta\right)}
}{
\cos^2{\left(\theta\right)}
}
\;\cdots\;1-\sin^2{\left(\theta\right)}=\cos^2{\left(\theta\right)},\;(A+B)^2=A^2+2AB+B^2
\\&=&
\frac{1}{\cos^2{\left(\theta\right)}}
+2\frac{\sin{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}
+\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}
\\&=&
\frac{1}{\cos^2{\left(\theta\right)}}
+2\frac{1}{\cos{\left(\theta\right)}}\frac{\sin{\left(\theta\right)}}{\cos{\left(\theta\right)}}
+\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}
\\&=&
\frac{1}{\cos^2{\left(\theta\right)}}
+2\frac{1}{\cos{\left(\theta\right)}}\tan{\left(\theta\right)}
+\tan^2{\left(\theta\right)}
\;\cdots\;\frac{\sin{\left(\theta\right)}}{\cos{\left(\theta\right)}}=\tan{\left(\theta\right)}
\\&=&
\left\{
\frac{1}{\cos{\left(\theta\right)}}
+\tan{\left(\theta\right)}
\right\}^2
\;\cdots\;A^2+2AB+B^2=(A+B)^2
\end{eqnarray}
$$
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