正弦関数のみを含む積分計算(1)

\begin{alignat}{2}
&(1) \displaystyle\int sin^2axdx=\frac{1}{2}x-\frac{1}{4a}sin2ax\\
&(2) \displaystyle\int sin^3axdx=-\frac{3}{4a}cosax+\frac{1}{12a}cos3ax\\
&(3) \displaystyle\int xsin^2axdx=\frac{1}{4}x^2-\frac{x}{4a}sin2ax-\frac{1}{8a^2}cos2ax\\
&(4) \displaystyle\int x^2sin^2axdx=\frac{1}{6}x^3-\left(\frac{x^2}{4a}-\frac{1}{8a^3}\right)sin2ax-\frac{x}{2a^2}cos2ax\\
&(5) \displaystyle\int sinaxsinbxdx=\frac{sin(a-b)x}{2(a-b)}-\frac{sin(a+b)x}{2(a+b)} (|a|≠|b|)\\
&(6) \displaystyle\int sin^naxdx=-\frac{sin^{n-1}axcosax}{na}+\frac{n-1}{n}\displaystyle\int sin^{n-2}axdx (n \gt 2)\\
&(7) \displaystyle\int \frac{1}{sinax}dx=\frac{1}{a}log\left|tan\frac{ax}{2}\right|\\
&(8) \displaystyle\int \frac{1}{sin^nax}dx=-\frac{cosax}{a(n-1)sin^{n-1}x}+\frac{n-2}{n-1}\displaystyle\int \frac{1}{sin^{n-2}x}dx (n \gt 1)\\
&(9) \displaystyle\int x^nsinaxdx=\displaystyle\sum_{k=0}^{2k \leq n}(-1)^{k+1}\frac{x^{n-2k}}{a^{2k+1}}
\cdot \frac{n!}{(n-2k)!}cosax\\
&                 +\displaystyle\sum_{k=0}^{2k+1 \leq n} (-1)^k\frac{x^{n-2k-1}}{a^{2k+2}}\cdot \frac{n!}{(n-2k-1)!}sinax
\end{alignat}(積分定数は省略しています。)





<証明>
(1) 半角の公式で次数を下げます。
\begin{alignat}{2}
&\displaystyle\int sin^2axdx=\frac{1}{2}\displaystyle\int (1-cos2ax)dx\\
&          =\frac{1}{2}\left(x-\frac{1}{2a}sin2ax\right)=\frac{1}{2}x-\frac{1}{4a}sin2ax
\end{alignat}


(2) 3倍角の公式を用いて次数を下げます。
\begin{alignat}{2}
&\displaystyle\int sin^3axdx=\displaystyle\int \left(\frac{3}{4}sinax-\frac{1}{4}sin3ax\right)dx\\
&          =-\frac{3}{4a}cosax+\frac{1}{12a}cos3ax
\end{alignat}


(3) 半角の公式で次数を下げて、部分積分です。
\begin{alignat}{2}
& \displaystyle\int xsin^2axdx=\frac{1}{2}\displaystyle\int x(1-cos2ax)dx\\
&           =\frac{1}{2}\displaystyle\int (x-xcos2ax)dx\\
&           =\frac{1}{2}\displaystyle\int xdx-\frac{1}{2}\displaystyle\int xcos2axdx\\
&           =\frac{1}{4}x^2-\frac{1}{2}\left(\frac{x}{2a}sin2ax-\displaystyle\int \frac{1}{2a}sin2axdx\right)\\
&           =\frac{1}{4}x^2-\frac{1}{2}\left(\frac{x}{2a}sin2ax+\frac{1}{4a^2}cos2axdx\right)\\
&           = \frac{1}{4}x^2-\frac{x}{4a}sin2ax-\frac{1}{8a^2}cos2ax
\end{alignat}


(4) 半角の公式で次数を下げて、部分積分です。
\begin{alignat}{2}
&\displaystyle\int x^2sin^2axdx=\frac{1}{2}\displaystyle\int x^2(1-cos2ax)dx\\
&=\frac{1}{2}\displaystyle\int x^2dx-\frac{1}{2}\displaystyle\int x^2cos2axdx\\
&=\frac{1}{6}x^3-\frac{1}{2}\left(\frac{x^2}{2a}sin2ax-\displaystyle\int \frac{2x}{2a}sin2axdx\right)\\
&=\frac{1}{6}x^3-\frac{x^2}{4a}sin2ax+\frac{1}{2a}\displaystyle\int xsin2axdx\\
&=\frac{1}{6}x^3-\frac{x^2}{4a}sin2ax+\frac{1}{2a}\left(-\frac{x}{2a}cos2ax+\displaystyle\int \frac{1}{2a}cos2axdx\right)\\
&=\frac{1}{6}x^3-\frac{x^2}{4a}sin2ax-\frac{x}{4a^2}cos2ax+\frac{1}{4a^2}\cdot \frac{1}{2a}sin2ax\\
&=\frac{1}{6}x^3-\left(\frac{x^2}{4a}-\frac{1}{8a^3}\right)sin2ax-\frac{x}{2a^2}cos2ax
\end{alignat}


(5) 積差の公式で被積分関数を切り離します。
\begin{alignat}{2}
& \displaystyle\int sinaxsinbxdx=-\frac{1}{2}\displaystyle\int \{cos(a+b)x-cos(a-b)x\}dx\\
&              =-\frac{1}{2}\cdot \frac{sin(a+b)x}{a+b}+\frac{1}{2}\cdot \frac{sin(a-b)x}{a-b}\\
&              =\frac{sin(a-b)x}{2(a-b)}-\frac{sin(a+b)x}{2(a+b)}
\end{alignat}


(6) \(sinax\) を一つだけ取り出して、部分積分です。
\begin{alignat}{2}
&\displaystyle\int sin^naxdx=\displaystyle\int sinaxsin^{n-1}axdx\\
&=-\frac{1}{a}cosaxsin^{n-1}ax-\displaystyle\int a(n-1)sin^{n-2}axcosax\left(-\frac{1}{a}cosax\right)dx\\
&=-\frac{sin^{n-1}axcosax}{a}+(n-1)\displaystyle\int sin^{n-2}axcos^2axdx\\
&=-\frac{sin^{n-1}axcosax}{a}+(n-1)\displaystyle\int sin^{n-2}ax(1-sin^2ax)dx\\
&=-\frac{sin^{n-1}axcosax}{a}+(n-1)\displaystyle\int sin^{n-2}ax-(n-1)\displaystyle\int sin^naxdx\\
\end{alignat}右の項(積分)を左辺に移項します。$$n\displaystyle\int sin^naxdx=-\frac{sin^{n-1}axcosax}{a}+(n-1)\displaystyle\int sin^{n-2}axdx$$両辺を \(n\) で割ります。$$\displaystyle\int sin^naxdx=-\frac{sin^{n-1}axcosax}{na}+\frac{n-1}{n}\displaystyle\int sin^{n-2}axdx$$


(7) \(\displaystyle tan\frac{ax}{2}=t\) と置きます。このとき
\begin{alignat}{2}
&sinax=2sin\frac{ax}{2}cos\frac{ax}{2}=2tan\frac{ax}{2}cos^2\frac{ax}{2}\\
&     =\frac{2tan\frac{ax}{2}}{1+tan^2\frac{ax}{2}}=\frac{2t}{1+t^2}\\
&\frac{a}{2}\cdot \frac{1}{cos^2\frac{ax}{2}}dx=dt, \frac{a}{2}(1+t^2)dx=dt, dx=\frac{2}{a(1+t^2)}dt
\end{alignat}となります。よって
\begin{alignat}{2}
&\displaystyle\int \frac{1}{sinax}dx=\displaystyle\int \frac{1+t^2}{2t}\cdot \frac{2}{a(1+t^2)}dt\\
&          =\frac{1}{a}\displaystyle\int \frac{1}{t}dt=\frac{1}{a}log|t|= \frac{1}{a}log\left|tan\frac{ax}{2}\right|
\end{alignat}


(8) \(\displaystyle \frac{1}{sin^2ax}\) を取り出して \(\displaystyle \frac{1}{sin^2ax}=-\frac{1}{a}\left(\frac{1}{tanax}\right)’ \) と見て、
  部分積分を行います。
\begin{alignat}{2}
&\displaystyle\int \frac{1}{sin^nax}dx=\displaystyle\int \frac{1}{sin^{n-2}axsin^2ax}dx\\
&=-\frac{1}{a}\displaystyle\int \frac{1}{sin^{n-2}ax}\cdot \left(\frac{1}{tanax}\right)’ dx\\
&=-\frac{1}{a}\left\{\frac{1}{sin^{n-2}axtanax}+\displaystyle\int a(n-2)sin^{-n+1}axcosax \cdot \frac{1}{tanax}dx\right\}\\
&=-\frac{1}{a}\left\{\frac{1}{sin^{n-2}axtanax}+a(n-2)\displaystyle\int \frac{1}{sin^{n-2}ax}\cdot \frac{1}{tan^2ax}dx\right\}\\
&=-\frac{1}{a}\left\{\frac{1}{sin^{n-2}axtanax}+a(n-2)\displaystyle\int \frac{1}{sin^{n-2}ax}\left(\frac{1}{sin^2ax}-1\right) dx\right\}\\
&=-\frac{1}{a}\left\{\frac{1}{sin^{n-2}axtanax}+a(n-2)\displaystyle\int \frac{1}{sin^nax}dx-a(n-2)\displaystyle\int \frac{1}{sin^{n-2}ax}dx\right\}\\
&=-\frac{cosax}{asin^{n-1}ax}-(n-2)\displaystyle\int \frac{1}{sin^nax}dx+(n-2)\displaystyle\int \frac{1}{sin^{n-2}ax}dx\\
\end{alignat}真ん中の項(積分)を左辺に移項します。$$ (n-1)\displaystyle\int \frac{1}{sin^nax}dx=-\frac{cosax}{asin^{n-1}ax}+(n-2)\displaystyle\int \frac{1}{sin^{n-2}ax}dx$$両辺を \(n-1\) で割ります。$$ \displaystyle\int \frac{1}{sin^nax}dx=-\frac{cosax}{a(n-1)sin^{n-1}x}+\frac{n-2}{n-1}\displaystyle\int \frac{1}{sin^{n-2}x}dx $$


(9) 部分積分を繰り返します。
\begin{alignat}{2}
&\displaystyle\int x^nsinaxdx=-\frac{x^n}{a}cosax+\displaystyle\int nx^{n-1}\frac{1}{a}cosaxdx\\
&           =-\frac{x^n}{a}cosax+\frac{n}{a}\displaystyle\int x^{n-1}cosaxdx\\
&\displaystyle\int x^{n-1}cosaxdx=\frac{x^{n-1}}{a}sinax-\displaystyle\int (n-1)x^{n-2}\frac{1}{a}sinaxdx\\
&             =\frac{x^{n-1}}{a}sinax-\frac{n-1}{a}\displaystyle\int x^{n-2}sinaxdx\\
&\displaystyle\int x^{n-2}sinaxdx=-\frac{x^{n-2}}{a}cosax+\displaystyle\int (n-2)x^{n-3}\frac{1}{a}cosaxdx\\
&             =-\frac{x^{n-2}}{a}cosax+\frac{n-2}{a}\displaystyle\int x^{n-3}cosaxdx\\
&\displaystyle\int x^{n-3}cosaxdx=\frac{x^{n-3}}{a}sinax-\displaystyle\int (n-3)x^{n-4}\frac{1}{a}sinaxdx\\
&             =\frac{x^{n-3}}{a}sinax-\frac{n-3}{a}\displaystyle\int x^{n-4}sinaxdx\\
&\\
&                \cdots
\end{alignat}
\((A)\) \(cosax\) について
\begin{alignat}{2}
&-\frac{x^n}{a}cosax+\frac{n}{a}\left(-\frac{n-1}{a}\right) \left(-\frac{x^{n-2}}{a}cosax\right)+\frac{n}{a}\left(-\frac{n-1}{a}\right)\left(\frac{n-2}{a}\right) \left(-\frac{n-3}{a}\right)\left(-\frac{x^{n-4}}{a}cosax\right)+ \cdots\\
&=-\frac{x^n}{a}cosax+\frac{n(n-1)x^{n-2}}{a^3}cosax-\frac{n(n-1)(n-2)(n-3)x^{n-4}}{a^5}cosax+ \cdots\\
&=-\frac{x^n}{a}cosax+\frac{x^{n-2}}{a^3}\cdot \frac{n!}{(n-2)!}cosax-\frac{x^{n-4}}{a^5}\cdot \frac{n!}{(n-4)!}cosax+ \cdots\\
&          \cdots +(-1)^{k+1}\frac{x^{n-2k}}{a^{2k+1}}\cdot \frac{n!}{(n-2k)!}cosax+ \cdots
\end{alignat}このとき \(2k\) は \(n\) を越えることはないので、次のように書けます。$$=\displaystyle\sum_{k=0}^{2k \leq n}(-1)^{k+1}\frac{x^{n-2k}}{a^{2k+1}}
\cdot \frac{n!}{(n-2k)!}cosax $$
\((B)\) \(sinax\) について
\begin{alignat}{2}
&\frac{n}{a}\cdot\frac{x^{n-1}}{a}sinax+\frac{n}{a}\left(-\frac{n-1}{a}\right) \left(\frac{n-2}{a}\right)\frac{x^{n-3}}{a}sinax+\frac{n}{a}\left(-\frac{n-1}{a}\right)\left(\frac{n-2}{a}\right) \left(-\frac{n-3}{a}\right)\left(\frac{n-4}{a}\right)\frac{x^{n-5}}{a}sinax+ \cdots\\
&=\frac{nx^{n-1}}{a^2}sinax-\frac{n(n-1)(n-2)x^{n-3}}{a^4}sinax+\frac{n(n-1)(n-2)(n-3)(n-4)x^{n-5}}{a^6}sinax+ \cdots\\
&=\frac{nx^{n-1}}{a^2}sinax-\frac{x^{n-3}}{a^4}\cdot \frac{n!}{(n-3)!}sinax+\frac{x^{n-5}}{a^6}\cdot \frac{n!}{(n-5)!}sinax+ \cdots\\
&          \cdots +(-1)^k\frac{x^{n-2k-1}}{a^{2k+2}}\cdot \frac{n!}{(n-2k-1)!}sinax+ \cdots
\end{alignat}このとき \(2k+1\) は \(n\) を越えることはないので、次のように書けます。 $$=\displaystyle\sum_{k=0}^{2k+1 \leq n} (-1)^k\frac{x^{n-2k-1}}{a^{2k+2}}\cdot \frac{n!}{(n-2k-1)!}sinax $$よって、次の結果を得ます。$$\displaystyle\int x^nsinaxdx=\displaystyle\sum_{k=0}^{2k \leq n}(-1)^{k+1}\frac{x^{n-2k}}{a^{2k+1}}
\cdot \frac{n!}{(n-2k)!}cosax+\displaystyle\sum_{k=0}^{2k+1 \leq n} (-1)^k\frac{x^{n-2k-1}}{a^{2k+2}}\cdot \frac{n!}{(n-2k-1)!}sinax$$

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