上級者
数学Ⅲ：積分法
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ヒント：\(\displaystyle \int \frac{1}{1+x^2}dx\)の形にするために，2乗の形に変換するのと1を作る必要がある．
問題
\(\displaystyle \lim_{N\rightarrow \infty}\int_{0}^{N} \frac{1}{1+x^2}dx=\frac{\pi}{2}\)を用いて，次の積分値を求めよ．
(A)\(\displaystyle I=\lim_{n\rightarrow \frac{\pi}{4}-0}\int_{0}^{n} \frac{1}{\sin^4 x+\cos^4 x}dx\)
(B)\(\displaystyle I=\lim_{n\rightarrow \frac{\pi}{4}-0}\int_{0}^{n} \frac{1}{\sin^6 x+\cos^6 x}dx\)
解答
(A)
\(\sin^4 x+\cos^4 x\)\(=(\cos^2 x-\sin^2 x)^2+2\sin^2 x\cos^2 x\)\(=\displaystyle \cos^2 2x+\frac{\sin^2 2x}{2}\)\(=\displaystyle \cos^2 2x\left(1+\frac{\tan^2 2x}{2}\right)\)
よって，\(I\)\(=\displaystyle \lim_{n\rightarrow \frac{\pi}{4}-0}\int_{0}^{n} \frac{\frac{1}{\cos^2 2x}}{1+\frac{\tan^2 2x}{2}}dx\)
ここで，\(\displaystyle \frac{\tan 2x}{\sqrt{2}}=t\)と置くと，\(\displaystyle dt=\frac{\sqrt{2}}{\cos^2 2x}dx\)，\(\displaystyle x:0\rightarrow n\)で\(\displaystyle t:0\rightarrow N\)
なお，\(\displaystyle N=\lim_{n\rightarrow \frac{\pi}{4}-0} \frac{\tan 2n}{\sqrt{2}}=\infty\)
よって，\(I\)\(=\displaystyle \lim_{N\rightarrow \infty}\frac{1}{\sqrt{2}}\int_{0}^{N}\frac{1}{1+t^2}dt\)\(=\displaystyle \frac{\pi}{2\sqrt{2}}\)
(B)
\(\sin^6 x+\cos^6 x\)\(=(\sin^2 x+\cos^2 x)^3-3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x)\)\(=1-3\sin^2 x\cos^2 x\)\(=\displaystyle (\sin^2 2x+\cos^2 2x)-\frac{3\sin^2 2x}{4}\)\(=\displaystyle \cos^2 2x+\frac{\sin^2 2x}{4}\)\(=\displaystyle \cos^2 2x\left(1+\frac{\tan^2 2x}{4}\right)\)
よって，\(I\)\(=\displaystyle \lim_{n\rightarrow \frac{\pi}{4}-0} \int_{0}^{n} \frac{\frac{1}{\cos^2 2x}}{1+\frac{\tan^2 2x}{4}}dx\)
ここで，\(\displaystyle \frac{\tan 2x}{2}=t\)と置くと，\(\displaystyle dt=\frac{1}{\cos^2 2x}dx\)，\(\displaystyle x:0\rightarrow n\)で\(\displaystyle t:0\rightarrow N\)
ここで，\(\displaystyle N=\lim_{n\rightarrow \frac{\pi}{4}-0} \frac{\tan 2n}{2}=\infty\)
よって，\(I\)\(=\displaystyle \lim_{N\rightarrow \infty} \int_{0}^{N} \frac{1}{1+t^2}dt\)\(=\displaystyle \frac{\pi}{2}\)