【微分積分】4-4-3 三角関数の積分｜要点まとめ
このページでは、大学数学の微分積分で扱う「三角関数の積分」について整理します。三角関数の基本公式や積分公式を確認し、例題を通して具体的な計算手順を理解します。公式の使い方や変形のコツを押さえ、積分計算力を養いましょう。
三角関数の積分
【三角関数の万能置換】
\(\displaystyle t=\tan\frac{x}{2}\)とおくと
\(\displaystyle \cos x=\frac{1-t^2}{1+t^2}\)
\(\displaystyle \sin x=\frac{2t}{1+t^2}\)
\(\displaystyle dx=\frac{2}{1+t^2}dt\)
が成り立つ。
\(f(u,v)\)を\(u\)と\(v\)に関する有理関数とする。このとき、\(\displaystyle t=\tan\frac{x}{2}\)とおくと
\(\displaystyle \int f(\cos x,\sin x)dx\)
は\(t\)に関する有理関数の積分で表すことができる。
【例題】次の不定積分・定積分を求めなさい。
(1)\(\displaystyle \int\frac{1}{2+\cos x}dx\)
\(\displaystyle t=\tan\frac{x}{2}\)とおくと
\(\displaystyle =\int\frac{1}{2+\frac{1-t^2}{1+t^2}}・\frac{2}{1+t^2}dt\)
\(\displaystyle =\int\frac{2}{3+t^2}dt\)
\(\displaystyle =\frac{2}{\sqrt{3}}\tan^{-1}\frac{t}{\sqrt{3}}+C\)
\(\displaystyle =\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{1}{\sqrt{3}}\tan\frac{x}{2}\right)+C\)
\(\displaystyle =\int\frac{1}{2+\frac{1-t^2}{1+t^2}}・\frac{2}{1+t^2}dt\)
\(\displaystyle =\int\frac{2}{3+t^2}dt\)
\(\displaystyle =\frac{2}{\sqrt{3}}\tan^{-1}\frac{t}{\sqrt{3}}+C\)
\(\displaystyle =\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{1}{\sqrt{3}}\tan\frac{x}{2}\right)+C\)
(2)\(\displaystyle \int\frac{1}{1+\sin x}dx\)
\(\displaystyle t=\tan\frac{x}{2}\)とおくと
\(\displaystyle =\int\frac{1}{1+\frac{2t}{1+t^2}}・\frac{2}{1+t^2}dt\)
\(\displaystyle =\int\frac{2}{(t+1)^2}dt\)
\(\displaystyle =-\frac{2}{t+1}+C\)
\(\displaystyle =-\frac{2}{1+\tan\frac{x}{2}}+C\)
\(\displaystyle =\int\frac{1}{1+\frac{2t}{1+t^2}}・\frac{2}{1+t^2}dt\)
\(\displaystyle =\int\frac{2}{(t+1)^2}dt\)
\(\displaystyle =-\frac{2}{t+1}+C\)
\(\displaystyle =-\frac{2}{1+\tan\frac{x}{2}}+C\)
(3)\(\displaystyle \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}}\frac{\cos x}{(1+\cos x)\sin x}dx\)
\(\displaystyle t=\tan\frac{x}{2}\)とおくと
\(\displaystyle =\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{\frac{1-t^2}{1+t^2}}{(1+\frac{1-t^2}{1+t^2})\frac{2t}{1+t^2}}・\frac{2}{1+t^2}dt\)
\(\displaystyle =\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\left(\frac{1}{2t}-\frac{t}{2}\right)dt\)
\(\displaystyle =\left[\frac{1}{2}\log|t|-\frac{t^2}{4}\right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\)
\(\displaystyle =\log\sqrt{3}-\frac{2}{3}\)
\(\displaystyle =\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{\frac{1-t^2}{1+t^2}}{(1+\frac{1-t^2}{1+t^2})\frac{2t}{1+t^2}}・\frac{2}{1+t^2}dt\)
\(\displaystyle =\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\left(\frac{1}{2t}-\frac{t}{2}\right)dt\)
\(\displaystyle =\left[\frac{1}{2}\log|t|-\frac{t^2}{4}\right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\)
\(\displaystyle =\log\sqrt{3}-\frac{2}{3}\)
(4)\(\displaystyle \int\frac{\cos x}{2+\sin^2x}dx\)
\(t=\sin x\)とおくと、\(\displaystyle dx=\frac{1}{\cos x}dt\)
\(\displaystyle =\int\frac{1}{t^2+2}dt\)
\(\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\frac{t}{\sqrt{2}}+C\)
\(\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\frac{\sin x}{\sqrt{2}}+C\)
\(\displaystyle =\int\frac{1}{t^2+2}dt\)
\(\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\frac{t}{\sqrt{2}}+C\)
\(\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\frac{\sin x}{\sqrt{2}}+C\)
(5)\(\displaystyle \int\frac{\cos x}{\cos x+\sin x}dx\)
\(\displaystyle =\int\frac{1}{1+\tan x}dx\)
\(t=\tan x\)とおくと、\(\displaystyle dx=\frac{1}{1+t^2}dt\)
\(\displaystyle =\frac{1}{2}\int\left(\frac{1}{1+t}+\frac{1-t}{1+t^2}\right)dt\)
\(\displaystyle =\frac{1}{4}\log\left|\frac{(1+t)^2}{1+t^2}\right|+\frac{1}{2}\tan^{-1}t+C\)
\(\displaystyle =\frac{1}{4}\log\left|\frac{(1+\tan x)^2}{1+\tan^2x}\right|+\frac{1}{2}\tan^{-1}(\tan x)+C\)
\(\displaystyle =\frac{1}{2}\log|\cos x+\sin x|+\frac{x}{2}+C\)
\(t=\tan x\)とおくと、\(\displaystyle dx=\frac{1}{1+t^2}dt\)
\(\displaystyle =\frac{1}{2}\int\left(\frac{1}{1+t}+\frac{1-t}{1+t^2}\right)dt\)
\(\displaystyle =\frac{1}{4}\log\left|\frac{(1+t)^2}{1+t^2}\right|+\frac{1}{2}\tan^{-1}t+C\)
\(\displaystyle =\frac{1}{4}\log\left|\frac{(1+\tan x)^2}{1+\tan^2x}\right|+\frac{1}{2}\tan^{-1}(\tan x)+C\)
\(\displaystyle =\frac{1}{2}\log|\cos x+\sin x|+\frac{x}{2}+C\)
(6)\(\displaystyle \int\frac{\tan x}{1+\cos^2x}dx\)
\(\displaystyle =\int\frac{\tan x}{1+\frac{1}{1+\tan^2x}}dx\)
\(t=\tan x\)とおくと、\(\displaystyle dx=\frac{1}{1+t^2}dt\)
\(\displaystyle =\int\frac{t}{1+\frac{1}{1+t^2}}・\frac{1}{1+t^2}dt\)
\(\displaystyle =\int\frac{t}{t^2+2}dt\)
\(\displaystyle =\frac{1}{2}\log(t^2+2)+C\)
\(\displaystyle =\frac{1}{2}\log(\tan^2x+2)+C\)
\(t=\tan x\)とおくと、\(\displaystyle dx=\frac{1}{1+t^2}dt\)
\(\displaystyle =\int\frac{t}{1+\frac{1}{1+t^2}}・\frac{1}{1+t^2}dt\)
\(\displaystyle =\int\frac{t}{t^2+2}dt\)
\(\displaystyle =\frac{1}{2}\log(t^2+2)+C\)
\(\displaystyle =\frac{1}{2}\log(\tan^2x+2)+C\)
次の学習に進もう！