1.次の不定積分を求めなさい。
(1)\(\displaystyle \int\frac{x^2+x}{x-2}dx\)
\(x^2+x=(x-2)(x+3)+6\)より、
\(\displaystyle =\int\frac{(x-2)(x+3)+6}{x-2}dx\)
\(\displaystyle =\int\left(x+3+\frac{6}{x-2}\right)dx\)
\(\displaystyle =\frac{1}{2}x^2+3x+6\log|x-2|+C\ \ \)(\(C\)は積分定数)
(2)\(\displaystyle \int\frac{x^2-1}{x+2}dx\)
\(x^2-1=(x+2)(x-2)+3\)より、
\(\displaystyle =\int\frac{(x+2)(x-2)+3}{x+2}dx\)
\(\displaystyle =\int\left(x-2+\frac{3}{x+2}\right)dx\)
\(\displaystyle =\frac{1}{2}x^2-2x+3\log|x+2|+C\ \ \)(\(C\)は積分定数)
(3)\(\displaystyle \int\frac{4x^2}{2x-1}dx\)
\(4x^2=(2x+1)(2x-1)+1\)より、
\(\displaystyle =\int\frac{(2x+1)(2x-1)+1}{2x-1}dx\)
\(\displaystyle =\int\left(2x+1+\frac{1}{2x-1}\right)dx\)
\(\displaystyle =x^2+x+\frac{1}{2}\log|2x-1|+C\ \ \)(\(C\)は積分定数)
(4)\(\displaystyle \int\frac{1}{x^2+x}dx\)
\(\displaystyle \frac{1}{x^2+x}=\frac{1}{x}-\frac{1}{x+1}\)より、
\(\displaystyle =\int\left(\frac{1}{x}-\frac{1}{x+1}\right)dx\)
\(\displaystyle =\log|x|-\log|x+1|+C\)
\(\displaystyle =\log\left|\frac{x}{x+1}\right|+C\ \ \)(\(C\)は積分定数)
(5)\(\displaystyle \int\frac{3}{x^2+x-2}dx\)
\(\displaystyle =\int\frac{3}{(x+2)(x-1)}dx\)
\(\displaystyle \frac{3}{(x+2)(x-1)}=\frac{1}{x-1}-\frac{1}{x+2}\)より、
\(\displaystyle =\int\left(\frac{1}{x-1}-\frac{1}{x+2}\right)dx\)
\(\displaystyle =\log|x-1|-\log|x+2|+C\)
\(\displaystyle =\log\left|\frac{x-1}{x+2}\right|+C\ \ \)(\(C\)は積分定数)
(6)\(\displaystyle \int\cos^3xdx\)
\(\displaystyle =\int(1-\sin^2x)\cos xdx\)
\(\sin x=t\)とおくと、\(\cos xdx=dt\)なので、
\(\displaystyle =\int(1-t^2)dt\)
\(\displaystyle =t-\frac{1}{3}t^3+C\)
\(\displaystyle =-\frac{1}{3}\sin^3x+\sin x+C\ \ \)(\(C\)は積分定数)
(7)\(\displaystyle \int\sin x\cos xdx\)
\(\displaystyle =\frac{1}{2}\int\sin2xdt\)
\(\displaystyle =-\frac{1}{4}\cos2x+C\ \ \)(\(C\)は積分定数)
(8)\(\displaystyle \int\sin^2xdx\)
\(\displaystyle \sin^2x=\frac{1-\cos2x}{2}\)より、
\(\displaystyle =\int\frac{1-\cos2x}{2}dx\)
\(\displaystyle =\frac{1}{2}x-\frac{1}{4}\sin2x+C\ \ \)(\(C\)は積分定数)
(9)\(\displaystyle \int\sin^23xdx\)
\(\displaystyle \sin^23x=\frac{1-\cos6x}{2}\)より、
\(\displaystyle =\int\frac{1-\cos6x}{2}dx\)
\(\displaystyle =\frac{1}{2}x-\frac{1}{12}\sin6x+C\ \ \)(\(C\)は積分定数)
(10)\(\displaystyle \int\cos^4xdx\)
\(\displaystyle \cos^4x=(\cos^2x)^2\)
\(\displaystyle =\left(\frac{1+\cos2x}{2}\right)^2\)
\(\displaystyle =\frac{1}{4}(1+2\cos2x+\frac{1+\cos4x}{2})\)
\(\displaystyle =\frac{1}{8}(3+4\cos2x+\cos4x)\)
よって、
\(\displaystyle \int\cos^4xdx=\frac{1}{8}\int(3+4\cos2x+\cos4x)dx\)
\(\displaystyle =\frac{3}{8}x+\frac{1}{4}\sin2x+\frac{1}{32}\sin4x+C\ \ \)(\(C\)は積分定数)
(11)\(\displaystyle \int\cos3x\cos2xdx\)
\(\displaystyle \cos\alpha\cos\beta=\frac{1}{2}\{\cos(\alpha+\beta)+\cos(\alpha-\beta)\}\)より、
\(\displaystyle =\frac{1}{2}\int\{\cos(3x+2x)+\cos(3x-2x)\}dx\)
\(\displaystyle =\frac{1}{2}\int(\cos5x+\cos x)dx\)
\(\displaystyle =\frac{1}{10}\sin5x+\frac{1}{2}\sin x+C\ \ \)(\(C\)は積分定数)
(12)\(\displaystyle \int\sin x\sin3xdx\)
\(\displaystyle \sin\alpha\sin\beta=\frac{1}{2}\{\cos(\alpha-\beta)-\cos(\alpha+\beta)\}\)より、
\(\displaystyle =\frac{1}{2}\int\{\cos(3x-x)-\cos(3x+x)\}dx\)
\(\displaystyle =\frac{1}{2}\int(\cos2x-\cos4x)dx\)
\(\displaystyle =\frac{1}{4}\sin2x-\frac{1}{8}\sin4x+C\ \ \)(\(C\)は積分定数)
(13)\(\displaystyle \int\sin3x\cos2xdx\)
\(\displaystyle \sin\alpha\cos\beta=\frac{1}{2}\{\sin(\alpha+\beta)+\sin(\alpha-\beta)\}\)より、
\(\displaystyle =\frac{1}{2}\int\{\sin(3x+2x)+\sin(3x-2x)\}dx\)
\(\displaystyle =\frac{1}{2}\int(\sin5x+\sin x)dx\)
\(\displaystyle =-\frac{1}{10}\cos5x-\frac{1}{2}\cos x+C\ \ \)(\(C\)は積分定数)
