【微分積分】4-4-3 三角関数の積分｜問題集

\(t=\sin x\)とおくと、\(\displaystyle dx=\frac{1}{\cos x}dt\)
\(\displaystyle =\int t^3dt\)
\(\displaystyle =\frac{t^4}{4}+C\)
\(\displaystyle =\frac{\sin^4x}{4}+C\)

\(t=\sin3x\)とおくと、\(\displaystyle dx=\frac{1}{3\cos3x}dt\)
\(\displaystyle =\int\frac{t^2}{3}dt\)
\(\displaystyle =\frac{t^3}{9}+C\)
\(\displaystyle =\frac{\sin^33x}{9}+C\)

\(\displaystyle =\int\frac{1+\cos2x}{2}dx\)
\(\displaystyle =\frac{x}{2}+\frac{1}{4}\sin2x+C\)

\(\displaystyle =\int\cos^2x\cos xdx\)
\(\displaystyle =\int(1-\sin^2x)\cos xdx\)
\(t=\sin x\)とおくと、\(\displaystyle dx=\frac{1}{\cos x}dt\)
\(\displaystyle =\int(1-t^2)dt\)
\(\displaystyle =t-\frac{t^3}{3}+C\)
\(\displaystyle =\sin x-\frac{\sin^3x}{3}+C\)

\(\displaystyle =\int\cos^4x\sin^2x\sin xdx\)
\(\displaystyle =\int\cos^4x(1-\cos^2x)\sin xdx\)
\(t=\cos x\)とおくと、\(\displaystyle dx=-\frac{1}{\sin x}dt\)
\(\displaystyle =\int-t^4(1-t^2)dt\)
\(\displaystyle =\frac{t^7}{7}-\frac{t^5}{5}+C\)
\(\displaystyle =\frac{\cos^7x}{7}-\frac{\cos^5x}{5}+C\)

\(\displaystyle =\frac{1}{2}\int(\sin5x-\sin x)dx\)
\(\displaystyle =-\frac{\cos5x}{10}+\frac{\cos x}{2}+C\)

\(\displaystyle =\frac{1}{2}\int(\cos x-\cos3x)dx\)
\(\displaystyle =\frac{\sin x}{2}-\frac{\sin3x}{6}+C\)

\(\displaystyle =\frac{1}{2}\int(\cos3x+\cos x)dx\)
\(\displaystyle =\frac{\sin3x}{6}+\frac{\sin x}{2}+C\)

\(t=\tan x\)とおくと、\(\displaystyle dx=\frac{1}{\sec^2x}dt\)
\(\displaystyle =\int tdt\)
\(\displaystyle =\frac{t^2}{2}+C\)
\(\displaystyle =\frac{\tan^2x}{2}+C\)

\(\displaystyle =\int\sec^2x\sec xdx\)
\(\displaystyle =\sec x\tan x-\int\sec x\tan^2xdx\)
\(\displaystyle =\sec x\tan x-\int\sec^3xdx+\int\sec xdx\)
\(\displaystyle I=\int\sec^3xdx\)とおくと
\(\displaystyle I=\sec x\tan x-I+\int\sec xdx\)
\(\displaystyle 2I=\sec x\tan x+\log|\sec x+\tan x|+C\)
\(\displaystyle I=\frac{\sec x\tan x}{2}+\frac{\log|\sec x+\tan x|}{2}+C\)
よって、
\(\displaystyle \int\sec^3xdx=\frac{\sec x\tan x}{2}+\frac{\log|\sec x+\tan x|}{2}+C\)

\(\displaystyle =\int\sin^2x\sin xdx\)
\(\displaystyle =\int(1-\cos^2x)\sin xdx\)
\(t=\cos x\)とおくと、\(\displaystyle dx=-\frac{1}{\sin x}dt\)
\(\displaystyle =\int-(1-t^2)dt\)
\(\displaystyle =\frac{t^3}{3}-t+C\)
\(\displaystyle =\frac{\cos^3x}{3}-\cos x+C\)

\(\displaystyle =\int\frac{1-\cos6x}{2}dx\)
\(\displaystyle =\frac{x}{2}-\frac{\sin6x}{12}+C\)

\(\displaystyle =\int\sin^2x\cos^2x\sin xdx\)
\(\displaystyle =\int(1-\cos^2x)\cos^2x\sin xdx\)
\(t=\cos x\)とおくと、\(\displaystyle dx=-\frac{1}{\sin x}dt\)
\(\displaystyle =\int-(1-t^2)t^2dt\)
\(\displaystyle =\frac{t^5}{5}-\frac{t^3}{3}+C\)
\(\displaystyle =\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C\)

\(\displaystyle =\frac{1}{2}\int(\sin5x-\sin x)dx\)
\(\displaystyle =-\frac{\cos5x}{10}+\frac{\cos x}{2}+C\)

\(\displaystyle =\int\sin^4x\sin xdx\)
\(\displaystyle =\int(1-\cos^2x)^2\sin xdx\)
\(t=\cos x\)とおくと、\(\displaystyle dx=-\frac{1}{\sin x}dt\)
\(\displaystyle =\int-(1-t^2)^2dt\)
\(\displaystyle =-\frac{t^5}{5}+\frac{2t^3}{3}-t+C\)
\(\displaystyle =-\frac{\cos^5x}{5}+\frac{2\cos^3x}{3}-\cos x+C\)

\(t=\pi x\)とおくと、\(\displaystyle dx=\frac{1}{\pi}dt\)
\(\displaystyle =\frac{1}{\pi}\int\sec^2tdt\)
\(\displaystyle =\frac{1}{\pi}\tan t+C\)
\(\displaystyle =\frac{1}{\pi}\tan\pi x+C\)

\(\displaystyle =\int\frac{(1-\cos^2x)\sin x}{\cos^3x}dx\)
\(t=\cos x\)とおくと、\(\displaystyle dx=-\frac{1}{\sin x}dt\)
\(\displaystyle =\int-\frac{1-t^2}{t^3}dt\)
\(\displaystyle =\int\left(\frac{1}{t}-t^{-3}\right)dt\)
\(\displaystyle =\log|t|+\frac{t^{-2}}{2}+C\)
\(\displaystyle =\log|\cos x|+\frac{1}{2}\sec^2 x+C\)

\(\displaystyle =\frac{1}{2}\tan^3x-\frac{1}{2}\int\tan^2x\sec^2xdx\)
\(\displaystyle I=\int\tan^2x\sec^2xdx\)とおくと
\(\displaystyle I=\frac{1}{2}\tan^3x-\frac{1}{2}I\)
\(\displaystyle 3I=\tan^3x+C\)
\(\displaystyle I=\frac{1}{3}\tan^3x+C\)
よって、
\(\displaystyle \int\tan^2x\sec^2xdx=\frac{1}{3}\tan^3x+C\)

\(\displaystyle =\int\tan^2x\sec^2x\sec x\tan xdx\)
\(\displaystyle =\int(\sec^2x-1)\sec^2x\sec x\tan xdx\)
\(t=\sec x\)とおくと、\(\displaystyle dx=\frac{1}{\sec x\tan x}dt\)
\(\displaystyle =\int(t^2-1)t^2dt\)
\(\displaystyle =\frac{t^5}{5}-\frac{t^3}{3}+C\)
\(\displaystyle =\frac{\sec^5x}{5}-\frac{\sec^3x}{3}+C\)

\(\displaystyle =\int\sec^3x\sec^2xdx\)
\(\displaystyle =\sec^3x\tan x-3\int\tan^2x\sec^3xdx\)
\(\displaystyle =\sec^3x\tan x-3\int\sec^5xdx+3\int\sec^3xdx\)
\(\displaystyle I=\int\sec^5xdx\)とおくと
\(\displaystyle I=\sec^3x\tan x-3I+3\int\sec^3xdx\)
\(\displaystyle 4I=\sec^3x\tan x+3\int\sec^3xdx\)
\(\displaystyle I=\frac{1}{4}\sec^3x\tan x+\frac{3}{4}\int\sec^3xdx\)
\(\displaystyle =\frac{1}{4}\sec^3x\tan x\)
\(\displaystyle \ \ \ +\frac{3}{8}(\sec x\tan x+\log|\sec x+\tan x|)+C\)
よって、
\(\displaystyle \int\sec^5xdx=\frac{1}{4}\sec^3x\tan x+\frac{3}{8}\sec x\tan x\)
\(\displaystyle \ \ \ +\frac{3}{8}\log|\sec x+\tan x|+C\)

\(\displaystyle t=\tan\frac{x}{2}\)とおくと
\(\displaystyle =\int\frac{1}{3-2\frac{1-t^2}{1+t^2}}・\frac{2}{1+t^2}dt\)
\(\displaystyle =\int\frac{2}{5t^2+1}dt\)
\(\displaystyle =\frac{2}{5}\int\frac{1}{t^2+\frac{1}{5}}dt\)
\(\displaystyle =\frac{2\sqrt{5}}{5}\tan^{-1}\sqrt{5}t+C\)
\(\displaystyle =\frac{2\sqrt{5}}{5}\tan^{-1}\left(\sqrt{5}\tan\frac{x}{2}\right)+C\)

\(\displaystyle t=\tan\frac{x}{2}\)とおくと
\(\displaystyle =\int\frac{\frac{2t}{1+t^2}}{2-\frac{2t}{1+t^2}}・\frac{2}{1+t^2}dt\)
\(\displaystyle =\int\frac{2t}{(1+t^2)(t^2-t+1)}dt\)
\(\displaystyle =\int\left(-\frac{2}{t^2+1}+\frac{2}{t^2-t+2}\right)dt\)
\(\displaystyle =-2\int\frac{1}{t^2+1}dt+\int\frac{2}{(t-\frac{1}{2})^2+\frac{3}{4}}dt\)
\(\displaystyle =-2\tan^{-1}t+\frac{4}{\sqrt{3}}\tan^{-1}\left(t-\frac{1}{2}\right)+C\)
\(\displaystyle =-2\tan^{-1}\tan\frac{x}{2}+\frac{4}{\sqrt{3}}\tan^{-1}\left(\tan\frac{x}{2}-\frac{1}{2}\right)+C\)
\(\displaystyle =-x+\frac{4}{\sqrt{3}}\tan^{-1}\left(\tan\frac{x}{2}-\frac{1}{2}\right)+C\)

\(\displaystyle t=\tan\frac{x}{2}\)とおくと
\(\displaystyle =\int\frac{1+\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}}・\frac{2}{1+t^2}dt\)
\(\displaystyle =\int\frac{2(t^2+2t+1)}{2(1+t^2)}dt\)
\(\displaystyle =\int\left(1+\frac{2t}{1+t^2}\right)dt\)
\(\displaystyle =t+\log|1+t^2|+C\)
\(\displaystyle =\tan\frac{x}{2}+\log\left|1+\tan^2\frac{x}{2}\right|+C\)

\(\displaystyle t=\tan\frac{x}{2}\)とおくと
\(\displaystyle =\int\frac{\frac{t^2}{1+t^2}}{\frac{t^2}{1+t^2}-\frac{1}{1+t^2}}・\frac{1}{1+t^2}dt\)
\(\displaystyle =\int\frac{t^2}{(t+1)(t-1)(1+t^2)}dt\)
\(\displaystyle =\frac{1}{4}\int\left(-\frac{1}{t+1}+\frac{1}{t-1}+\frac{2}{1+t^2}\right)dt\)
\(\displaystyle =-\frac{1}{4}\log|t+1|+\frac{1}{4}\log|t-1|+\frac{1}{2}\tan^{-1}t+C\)
\(\displaystyle =\frac{1}{4}\log\left|\frac{\tan\frac{x}{2}-1}{\tan\frac{x}{2}+1}\right|+\frac{x}{2}+C\)

\(\displaystyle t=\tan x\)とおくと
\(\displaystyle =\int\frac{1}{1+t}・\frac{1}{1+t^2}dt\)
\(\displaystyle =\int\left(\frac{1}{2(1+t)}+\frac{1-t}{2(1+t^2)}\right)dt\)
\(\displaystyle =\frac{1}{4}\log\left|\frac{(1+t)^2}{1+t^2}\right|+\frac{1}{2}\tan^{-1}t+C\)
\(\displaystyle =\frac{1}{2}\log|\cos x+\sin x|+\frac{x}{2}+C\)