三角関数の積分②(分数型)

引き続き三角関数の積分です。
今回は主に分数型のものを扱います。

(例題1)
次の不定積分を求めよ。
(1-1)\(\displaystyle\int\displaystyle\frac{1}{\cos x}dx\)
(1-2)\(\displaystyle\int\displaystyle\frac{1}{\cos^2 x}dx\)
(1-3)\(\displaystyle\int\displaystyle\frac{1}{\cos^3x}dx\)
(1-4)\(\displaystyle\int\displaystyle\frac{1}{\cos^4 x}dx\)

(2-1)\(\displaystyle\int\displaystyle\frac{1}{\tan x}dx\)
(2-2)\(\displaystyle\int\displaystyle\frac{1}{\tan^2 x}dx\)
(2-3)\(\displaystyle\int\displaystyle\frac{1}{\tan^3x}dx\)

(解答)
以下積分定数を\(C\)とする。
(1-1)
\(\displaystyle\int\displaystyle\frac{1}{\cos x}dx\)

\(=\displaystyle\int\displaystyle\frac{\cos x}{1-\sin^2x}dx\)

(\(\sin x=t\) と置換して)

\(=\displaystyle\int\displaystyle\frac{1}{1-t^2}dt\)

\(=\displaystyle\frac{1}{2}\displaystyle\int(\displaystyle\frac{1}{1-t}+\displaystyle\frac{1}{1+t})dt\)

\(=\displaystyle\frac{1}{2}(-\log|1-t|+\log|1+t|)+C\)

\(=\displaystyle\frac{1}{2}\log\left|\displaystyle\frac{1+t}{1-t}\right|+C\)

\(=\displaystyle\frac{1}{2}\log\left(\displaystyle\frac{1+\sin x}{1-\sin x}\right)+C\)

(1-2)
\(\displaystyle\int\displaystyle\frac{1}{\cos^2 x}dx\)

\(=\tan x+C\)

(1-3)
\(\displaystyle\int\displaystyle\frac{1}{\cos^3x}dx\)

\(=\displaystyle\int\displaystyle\frac{\cos x}{(1-\sin^2x)^2}dx\)

(\(\sin x=t\) と置換して)

\(=\displaystyle\int\displaystyle\frac{1}{(1-t^2)^2}dt\)

\(=\displaystyle\int\displaystyle\frac{1}{(1-t)^2(1+t)^2}dt\)・・・①

(部分分数分解します)
ここで
\(\displaystyle\int\displaystyle\frac{1}{(1-t)^2(1+t)^2}=\displaystyle\frac{A}{1-t}+\displaystyle\frac{B}{(1-t)^2}+\displaystyle\frac{C}{1+t}+\displaystyle\frac{D}{(1+t)^2}\)
とおいて、\(A,B,C,D\) を決定すると
\(A=B=C=D=\displaystyle\frac{1}{4}\)
となるから

①は
\(\displaystyle\frac{1}{4}\displaystyle\int\left\{\displaystyle\frac{1}{1-t}+\displaystyle\frac{1}{(1-t)^2}+\displaystyle\frac{1}{1+t}+\displaystyle\frac{1}{(1+t)^2}\right\}dt\)

\(=\displaystyle\frac{1}{4}\left\{-\log|1-t|+\displaystyle\frac{(-1)(-1)}{1-t}+\log|1+t|-\displaystyle\frac{1}{1+t}\right\}+C\)

\(=\displaystyle\frac{1}{4}\left\{\log\left|\displaystyle\frac{1+t}{1-t}\right|+\displaystyle\frac{2t}{1-t^2}\right\}+C\)

(変数をもとに戻して整理すると)

\(=\displaystyle\frac{1}{4}\left\{\log\left(\displaystyle\frac{1+\sin x}{1-\sin x}\right)+\displaystyle\frac{2\sin x}{\cos^2x}\right\}+C\)

(1-4)
\(\displaystyle\int\displaystyle\frac{1}{\cos^4 x}dx\)

\(=\displaystyle\int(\tan^2x+1)\displaystyle\frac{1}{\cos^2 x}dx\)

(導関数接触型、\(t^2+1\) の積分をして)

\(=\displaystyle\frac{1}{3}\tan^3x+\tan x+C\)

(2-1)
\(\displaystyle\int\displaystyle\frac{1}{\tan x}dx\)

\(=\displaystyle\int\displaystyle\frac{\cos x}{\sin x}dx\)

(分母を微分すると分子)

\(=\log|\sin x|+C\)

(2-2)
\(\displaystyle\int\displaystyle\frac{1}{\tan^2 x}dx\)

\(=\displaystyle\int(\displaystyle\frac{1}{\sin^2 x}-1)dx\)

\(=-\displaystyle\frac{1}{\tan x}-x+C\)

(2-3)
\(\displaystyle\int\displaystyle\frac{1}{\tan^3x}dx\)

\(=\displaystyle\int\displaystyle\frac{1}{\tan x}\cdot(\displaystyle\frac{1}{\sin^2 x}-1)dx\)

\(=\displaystyle\int(-\displaystyle\frac{1}{\tan x}\cdot\displaystyle\frac{-1}{\sin^2 x}-\displaystyle\frac{1}{\tan x})dx\)

(1項目は導関数接触型、\((\displaystyle\frac{1}{\tan x}=)\ t\) (1次式) の積分をする。2項目は(2-1)と同じ)

\(=-\displaystyle\frac{1}{2\tan^2x}-\log|\sin x|+C\)

(例題2)
次の不定積分を求めよ。
(1)\(\displaystyle\int\displaystyle\frac{1-\tan x}{1+\tan x}dx\)

(2-1)\(\displaystyle\int\displaystyle\frac{1}{1+\cos x}dx\)
(2-2)\(\displaystyle\int\displaystyle\frac{1}{1+\sin x}dx\)

(解答)
(1)
(\(\sin x,\cos x\)に変換すると)
\(\displaystyle\int\displaystyle\frac{1-\tan x}{1+\tan x}dx\)

\(=\displaystyle\int\displaystyle\frac{\cos x-\sin x}{\cos x+\sin x}dx\)

(分母を微分すると分子になるから)

\(=\log|\cos x+\sin x|+C\)

(2-1)

\(\displaystyle\int\displaystyle\frac{1}{1+\cos x}dx\)

(\(1-\cos x\) を分母分子に掛けて)

\(=\displaystyle\int\displaystyle\frac{1-\cos x}{\sin^2 x}dx\)

\(=\displaystyle\int(\displaystyle\frac{1}{\sin^2 x}-\displaystyle\frac{\cos x}{\sin^2x})dx\)

(2項目は導関数接触型)

\(=-\displaystyle\frac{1}{\tan x}+\displaystyle\frac{1}{\sin x}+C\)

(別解)
(半角(倍角)の公式を利用すると)
\(\displaystyle\int\displaystyle\frac{1}{1+\cos x}dx\)

\(=\displaystyle\int\displaystyle\frac{1}{2\cos^2\displaystyle\frac{x}{2}}dx\)

\(=\displaystyle\frac{1}{2}\cdot(\tan\displaystyle\frac{x}{2})\cdot2+C\)

\(=\tan\displaystyle\frac{x}{2}+C\)

(なお本解答の結果を変形してみると)
\(-\displaystyle\frac{1}{\tan x}+\displaystyle\frac{1}{\sin x}+C\)

\(=\displaystyle\frac{1-\cos x}{\sin  x}+C\)

\(=\displaystyle\frac{2\sin^2\displaystyle\frac{x}{2}}{2\sin\displaystyle\frac{x}{2}\cos\displaystyle\frac{x}{2}}+C\)

\(=\tan\displaystyle\frac{x}{2}+C\)
となり一致する。

(2-2)
\(\displaystyle\int\displaystyle\frac{1}{1+\sin x}dx\)

(\(1-\sin x\)を分母分子に掛けて)

\(=\displaystyle\int\displaystyle\frac{1-\sin x}{\cos^2x}dx\)

\(=\displaystyle\int(\displaystyle\frac{1}{\cos^2x}+\displaystyle\frac{-\sin x}{\cos^2x})dx\)

(2項目は導関数接触)

\(=\tan x-\displaystyle\frac{1}{\cos x}+C\)

(別解)

\(\displaystyle\int\displaystyle\frac{1}{1+\sin x}dx\)

\(=\displaystyle\int\displaystyle\frac{1}{1+\cos(\displaystyle\frac{π}{2}-x)}dx\)

\(=\displaystyle\int\displaystyle\frac{1}{2\cos^2(\displaystyle\frac{π}{4}-\displaystyle\frac{x}{2})}dx\)

\(=\displaystyle\frac{1}{2}\{\tan(\displaystyle\frac{π}{4}-\displaystyle\frac{x}{2})\}\cdot(-2)+C\)

\(=-\tan(\displaystyle\frac{π}{4}-\displaystyle\frac{x}{2})+C\)

(答えが一致することを確かめると)
\(-\tan(\displaystyle\frac{π}{4}-\displaystyle\frac{x}{2})\)

\(=-\displaystyle\frac{1-\tan\displaystyle\frac{x}{2}}{1+1\cdot\tan\displaystyle\frac{x}{2}}\)

\(=-\displaystyle\frac{1-\tan\displaystyle\frac{x}{2}}{1+\tan\displaystyle\frac{x}{2}}\)

\(=-\displaystyle\frac{\cos\displaystyle\frac{x}{2}-\sin\displaystyle\frac{x}{2}}{\cos\displaystyle\frac{x}{2}+\sin\displaystyle\frac{x}{2}}\)

\(=-\displaystyle\frac{(\cos\displaystyle\frac{x}{2}-\sin\displaystyle\frac{x}{2})^2}{\cos^2\displaystyle\frac{x}{2}-\sin^2\displaystyle\frac{x}{2}}\)

\(=-\displaystyle\frac{1-2\sin\displaystyle\frac{x}{2}\cos\displaystyle\frac{x}{2}}{\cos x}\)

\(=-\displaystyle\frac{1-\sin x}{\cos x}\)

\(=\tan x-\displaystyle\frac{1}{\cos x}\)

以上になります。お疲れさまでした。
ここまで見て頂きありがとうございました。
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