MATH

ヒント：難しい積分は解き方を覚えてないとなかなか解けない．今回は3つのパターンがあるので覚えておきましょう．

問題

定積分 \(I=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan x}dx\) を求めよ．（オリジナル）

三角関数の積分は，\(\displaystyle \tan \frac{x}{2}=t\)と置くと解けることが多いが，ほかの方法で解けるか確認しよう．

パターン１：

\(\displaystyle \tan \frac{x}{2}=t\)と置くと，\(\displaystyle \tan x=\frac{2t}{1-t^2}\)

\(\displaystyle x:0\rightarrow\frac{\pi}{2}\)の時，\(\displaystyle t:0\rightarrow 1\)

また，\(\displaystyle dx=\frac{2}{1+t^2}dt\)

よって，\(I=\displaystyle \int_{0}^{1}\frac{2(1-t^2)}{(1+t^2)(1-t^2+2t)}dt\)

ここで部分分数分解すると，

\(I=\displaystyle \int_{0}^{1}\left(\frac{1}{1+t^2}-\frac{t}{1+t^2}+\frac{t-1}{t^2-2t-1}\right)dt\)

ここで，

\(\displaystyle \int_{0}^{1}\frac{1}{1+t^2}dt\)について，\(t=\tan\theta\)と置いて計算すれば，

\(\displaystyle \int_{0}^{1}\frac{1}{1+t^2}dt\)\(=\displaystyle \frac{\pi}{4}\)

また，\(\displaystyle \int_{0}^{1}\frac{t}{1+t^2}\)\(=\displaystyle \int_{0}^{1}\frac{1}{2}\frac{(t^2+1)'}{1+t^2}\)\(=\displaystyle [\log(t^2+1)]_{0}^{1}\)\(=\displaystyle \frac{\log 2}{2}\)

また，\(\displaystyle \int_{0}^{1}\left(\frac{t-1}{t^2-2t-1}\right)dt\)\(=\displaystyle \int_{0}^{1}\left(\frac{1}{2}\frac{(t^2-2t-1)'}{t^2-2t-1}\right)dt\)\(=\displaystyle \left[\frac{1}{2}\log|t^2-2t-1|\right]_{0}^{1}\)\(=\displaystyle \frac{\log 2}{2}\)

よって，\(I=\displaystyle \frac{\pi}{4}\)

パターン２：

\(I=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan x}dx\)\(=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{\sin x+\cos x}dx\)

ここで，\(x=\displaystyle \frac{\pi}{2}-t\)と置くと，\(\displaystyle \sin x=\sin\left(\frac{\pi}{2}-t\right)=\cos t\)，\(\displaystyle \cos x=\cos\left(\frac{\pi}{2}-t\right)=\sin t\)

\(\displaystyle x:0\rightarrow\frac{\pi}{2}\)の時，\(\displaystyle t:\frac{\pi}{2}\rightarrow 0\)

また，\(dx=-dt\)

よって，\(I=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sin x+\cos x}dx\) (\(t\)や\(x\)は単なる積分変数なので変更可能)

ここで，\(2I=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{\sin x+\cos x}dx\)\(+\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sin x+\cos x}dx\)\(=\displaystyle \int_{0}^{\frac{\pi}{2}}dx\)\(=\displaystyle \frac{\pi}{2}\)

よって，\(I=\displaystyle \frac{\pi}{4}\)

パターン３：

\(I=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan x}dx\)\(=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{\sin x+\cos x}dx\)

ここで，

\(\displaystyle \sin x+\cos x=\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)\)

\(\displaystyle \cos x=\cos\left(x-\frac{\pi}{4}+\frac{\pi}{4}\right)\)\(=\displaystyle \cos\left(x-\frac{\pi}{4}\right)\cos \left(\frac{\pi}{4}\right)+\sin\left(x-\frac{\pi}{4}\right)\sin \left(\frac{\pi}{4}\right)\)\(=\displaystyle \frac{1}{\sqrt{2}}\cos\left(x-\frac{\pi}{4}\right)+ \frac{1}{\sqrt{2}}\sin\left(x-\frac{\pi}{4}\right)\)

よって，\(I=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan x}dx\)\(=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{\sin x+\cos x}dx\)\(=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{2}+\frac{1}{2}\tan\left(x-\frac{\pi}{4}\right)dx\)

ここで，\(\displaystyle \int \tan x dx\)\(=\displaystyle \int \frac{(-\cos x)'}{\cos x}dx\)\(=-\log|\cos x|\)

よって，\(I=\displaystyle \left[\frac{1}{2}-\frac{1}{2}\log\left|\cos \left(x-\frac{\pi}{4}\right)\right|\right]_{0}^{\frac{\pi}{2}}\)\(\displaystyle \frac{\pi}{4}\)

参考入試問題

解答