[数Ⅲ]不定積分の計算（sin x の積分）
問題 [目標10分]
(1) \(\displaystyle\int\sin{x}dx\)
(2) \(\displaystyle\int\sin^2{x}dx\)
(3) \(\displaystyle\int\sin^3{x}dx\)
(4) \(\displaystyle\int\sin^4{x}dx\)
2乗，3乗，4乗で計算の仕方が異なるよ．確認しておこう
【解答】\(C\) は積分定数
(1) \(\displaystyle\int\sin{x}dx＝－\cos{x}+C\)
(2) \(\displaystyle\int\sin^2{x}dx\)
\(\displaystyle＝\int\frac{1}{2}(1－\cos{2x})dx\)
\(\displaystyle＝\frac{1}{2}(x－\frac{1}{2}\sin{2x})＋C\)
\(\displaystyle＝\frac{1}{2}x－\frac{1}{4}\sin{2x}＋C\)
(3) \(\displaystyle\int\sin^3{x}dx\)
\(\displaystyle＝\int\sin^2{x}・\sin{x}dx\)
\(\displaystyle＝\int(1－\cos^2{x})・(－\cos{x})’dx\)
\(\displaystyle＝\int\lbrace(－\cos{x})’+\cos^2{x}(\cos{x})’\rbrace dx\)
\(\displaystyle＝－\cos{x}+\frac{1}{3}\cos^3{x}＋C\)
(4) \(\displaystyle\int\sin^4{x}dx\)
\(\displaystyle＝\int(\sin^2{x})^2dx\)
\(\displaystyle＝\int\lbrace\frac{1}{2}(1－\cos{2x})\rbrace ^2dx\)
ここで括弧の中を展開します
\(\displaystyle＝\int\frac{1}{4}(1－2\cos{2x}+\cos^2{2x})dx\)
\(\displaystyle＝\int\frac{1}{4}(1－2\cos{2x}+\frac{1+\cos{4x}}{2})dx\)
\(\displaystyle＝\frac{1}{4}\lbrace x－\sin{2x}+\frac{1}{2}(x+\frac{1}{4}\sin{4x})+C\rbrace\)
積分定数に数をかけても積分定数は\(C\) のままです
\(\displaystyle＝\frac{3}{8}x－\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+C\)
POINT
⇒ 公式利用
(2) \(\displaystyle\int\sin^2{x}dx\)
⇒\(\displaystyle\sin^2{x}＝\frac{1}{2}(1－\cos{2x})\)の利用
(3) \(\displaystyle\int\sin^3{x}dx\)
⇒\(\displaystyle\sin^2{x}・\sin{x}＝(1－\cos^2{x})・(－\cos{x})’\)の式変形
(4) \(\displaystyle\int\sin^4{x}dx\)
⇒\(\displaystyle(\sin^2{x})^2＝\lbrace\frac{1}{2}(1－\cos{2x})\rbrace ^2\)の利用
すなわち, (3) の3 乗だけ他と解き方が異なります。
