不定積分の問題

■問題

次の問題を積分せよ（不定積分）．

${\displaystyle {\displaystyle \int \sin 2x\sin xdx}}$ 　　

■答

${\displaystyle -\frac{1}{6}\sin 3x+\frac{1}{2}\sin x+C}$，または，${\displaystyle \frac{2}{3}{\sin }^{3}x+C}$　　（$C$は積分定数）

■ヒント

三角関数の積和の公式より

${\displaystyle \sin \alpha \sin \beta }$${\displaystyle =-\frac{1}{2}\left\{\cos \left(\alpha +\beta \right)-\cos \left(\alpha -\beta \right)\right\}}$ 　･･････${\displaystyle \left(1\right)}$

の公式を用いる．

■解説

この問題では，ヒントの公式 ${\displaystyle \left(1\right)}$ の ${\displaystyle \alpha }$ に${\displaystyle 2x}$ ，${\displaystyle \beta }$ に${\displaystyle x}$ をあてはめる．

与式${\displaystyle ={\displaystyle \int \left(-\frac{1}{2}\cos 3x+\frac{1}{2}\cos x\right)dx}}$

${\displaystyle =-\frac{1}{2}\cdot \frac{1}{3}\sin 3x+\frac{1}{2}\sin x+C}$

${\displaystyle =-\frac{1}{6}\sin 3x+\frac{1}{2}\sin x+C}$

（${\displaystyle C}$ は積分定数）

■別解　その1

${\displaystyle \sin 2x=2\sin x\cos x}$ より

与式${\displaystyle ={\displaystyle \int 2\sin x\cos x\sin xdx}}$

${\displaystyle =2{\displaystyle \int {\sin }^{2}x\cos xdx}}$

${\displaystyle =2{\displaystyle \int {\sin }^{2}x\cos xdx}}$

${\displaystyle \sin x=t}$とおく

${\displaystyle \frac{dt}{dx}=\cos x}$　→　${\displaystyle \cos xdx=dt}$

よって

${\displaystyle =2{\displaystyle \int t^{2}dt}}$

${\displaystyle =2\cdot \frac{1}{3}{t}^3+C}$

${\displaystyle =\frac{2}{3}{\sin }^{3}x+C}$

（${\displaystyle C}$ は積分定数）

■別解　その2

与式${\displaystyle ={\displaystyle \int 2\sin x\cos x\sin xdx}}$

${\displaystyle ={{\displaystyle \int \left(-\frac{1}{2}\cos 2x\right)}}^{\prime }\sin xdx}$

${\displaystyle =-\frac{1}{2}\cos 2x\sin x}$${\displaystyle -{\displaystyle \int \left(-\frac{1}{2}\cos 2x\right)}\cos xdx}$

${\displaystyle =-\frac{1}{2}\cos 2x\sin x}$${\displaystyle +\frac{1}{2}{\displaystyle \int \cos 2x\cos x}dx}$

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

${\displaystyle =-\frac{1}{2}\cos 2x\sin x}$${\displaystyle +\frac{1}{4}\sin 2x\cos x}$${\displaystyle +\frac{1}{4}{\displaystyle \int \sin 2x\left(-\sin x\right)}dx}$

よって

${\displaystyle \left(1-\frac{1}{4}\right){\displaystyle \int \sin 2x\sin xdx}}$${\displaystyle =-\frac{1}{2}\cos 2x\sin x+\frac{1}{4}\sin 2x\cos x}$

${\displaystyle =-\frac{1}{3}\left(-2\cos 2x\sin x+\sin 2x\cos x\right)}$

${\displaystyle =-\frac{2}{3}{\sin }^{3}x}$

最後に積分定数${\displaystyle C}$ を加え

${\displaystyle {\displaystyle \int \sin 2x\sin xdx=\frac{2}{3}{\sin }^{3}x+C}}$

■${\displaystyle -\frac{1}{6}\sin 3x+\frac{1}{2}\sin x+C}$ と${\displaystyle \frac{2}{3}{\sin }^{3}x+C}$が同じ理由

${\displaystyle \sin 3x}$

${\displaystyle =\sin \left(2x+x\right)}$

${\displaystyle =\sin 2x\cos x+\cos 2x\sin x}$

${\displaystyle =2\sin x\cos x\cos x}$${\displaystyle +\left({\cos }^{2}x-{\sin }^{2}x\right)\sin x}$

${\displaystyle =2\sin x\left(1-{\sin }^{2}x\right)}$${\displaystyle +\left(1-2{\sin }^{2}x\right)\sin x}$

${\displaystyle =2\sin x-2{\sin }^{3}x+\sin x-2{\sin }^{3}x}$

${\displaystyle =3\sin x-4{\sin }^{3}x}$

よって

${\displaystyle -\frac{1}{6}\sin 3x+\frac{1}{2}\sin x+C}$ ${\displaystyle =-\frac{1}{6}\left(3\sin x-4{\sin }^{3}x\right)+\frac{1}{2}\sin x+C}$

${\displaystyle =-\frac{1}{2}\sin x+\frac{2}{3}{\sin }^{3}x+\frac{1}{2}\sin x+C}$

${\displaystyle =\frac{2}{3}{\sin }^{3}x+C}$

　

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最終更新日： 2023年11月24日