tanの立方根の積分
tanの立方根の積分
\[ \int\sqrt[3]{\tan x}dx=\frac{1}{4}\log\left(\tan^{\frac{4}{3}}x-\tan^{\frac{2}{3}}x+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2\tan^{\frac{2}{3}}x-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left(\tan^{\frac{2}{3}}x+1\right)+C \]
\[ \int\sqrt[3]{\tan x}dx=\frac{1}{4}\log\left(\tan^{\frac{4}{3}}x-\tan^{\frac{2}{3}}x+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2\tan^{\frac{2}{3}}x-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left(\tan^{\frac{2}{3}}x+1\right)+C \]
\begin{align*}
\int\sqrt[3]{\tan x}dx & =3\int\frac{t^{3}}{t^{6}+1}dt\cmt{t=\sqrt[3]{\tan x}}\\
& =\frac{3}{2}\int\frac{u}{u^{3}+1}du\cmt{u=t^{2}}\\
& =\frac{1}{2}\int\left(\frac{u+1}{u^{2}-u+1}-\frac{1}{u+1}\right)du\\
& =\frac{1}{2}\int\left(\frac{\frac{1}{2}(u^{2}-u+1)'}{u^{2}-u+1}+\frac{\frac{3}{2}}{u^{2}-u+1}-\frac{1}{u+1}\right)du\\
& =\frac{1}{2}\int\left(\frac{\frac{1}{2}(u^{2}-u+1)'}{u^{2}-u+1}+\frac{\frac{3}{2}}{\left(u-\frac{1}{2}\right)^{2}+\frac{3}{4}}-\frac{1}{u+1}\right)du\\
& =\frac{1}{4}\log\left(u^{2}-u+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2u-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left|u+1\right|+C\\
& =\frac{1}{4}\log\left(\tan^{\frac{4}{3}}x-\tan^{\frac{2}{3}}x+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2\tan^{\frac{2}{3}}x-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left(\tan^{\frac{2}{3}}x+1\right)+C
\end{align*}
ページ情報
タイトル | tanの立方根の積分 |
URL | https://www.nomuramath.com/rnqo16ad/ |
SNSボタン |
分母分子に3角関数を含む定積分
\[
\int_{0}^{\frac{\pi}{2}}\frac{\sqrt[3]{\tan x}}{\left(\sin x+\cos x\right)^{2}}dx=?
\]
分母に双曲線関数で分子に3角関数の定積分
\[
\int_{-\infty}^{\infty}\frac{\cos x}{\cosh x}dx=?
\]
対数同士の積の積分
\[
\int_{0}^{1}\log\left(x\right)\log\left(1+x\right)dx=?
\]
指数関数を分母と分子に含む対数の定積分
\[
\int_{0}^{\infty}\log\left(\frac{e^{x}-1}{e^{x}+1}\right)dx=?
\]