삼각함수의 유리식 적분
Question.
- \(\int \frac{1}{a+\cos\theta}d\theta\)
- \(\int \frac{1}{a+\sin\theta}d\theta\)
IDEA
반각치환 (바이어-슈트라스 치환) 을 사용하자.
- We can take \(x=\tan\frac{\theta}{2} \rightarrow dx=\frac{1}{2}\sec^2\frac{\theta}{2}d\theta\). Then we get
- \(\cos\frac{\theta}{2}=\frac{x}{\sqrt{x^2+1}}\)
- \(\sin\frac{\theta}{2}=\frac{1}{\sqrt{x^2+1}}\)
- 반각공식을 사용하면, \(\cos\)에 대하여 \(\theta\)를 \(\frac{\theta}{2}\)로 변환할 수 있으므로, 이후 반각치환을 사용한다.
- \(\sin\)의 경우, 평행이동을 사용한다. \(\theta = \theta-\frac{\pi}{2}\)
Solution
As a result, without loss of generality, we need to solve the \(\int \frac{1}{a+\cos\theta}d\theta\).:
\[ \begin{aligned} \int \frac{1}{a+\cos\theta}d\theta &= \int\frac{1}{(a-1)+2\cos^2\frac{\theta}{2}}d\theta \\ &= \int\frac{1}{(a-1)+2\frac{x^2}{x^2+1}}\times2\frac{x^2}{x^2+1}dx = 2\int\frac{x^2}{(a-1)(x^2+1)+2x^2}dx=2\int\frac{x^2}{(a+1)x^2+(a-1)}dx = \frac{2}{a+1}\int\frac{x^2}{x^2+\frac{a-1}{a+1}}dx \\ &= \frac{2}{a+1}\int\left[1-\frac{a-1}{a+1}\frac{1}{x^2+\frac{a-1}{a+1}}\right]dx \\ &= \frac{2}{a+1}\left[x- \sqrt{\frac{a-1}{a+1}}\arctan\frac{x}{\sqrt{\frac{a-1}{a+1}}} \right]+C = \frac{2}{a+1}\left[\tan\frac{\theta}{2}- \sqrt{\frac{a-1}{a+1}}\arctan\frac{\tan\frac{\theta}{2}}{\sqrt{\frac{a-1}{a+1}}} \right]+C \end{aligned} \]
Moreover, we can solve the \(\int \frac{1}{a+\sin\theta}d\theta\) from above result.:
\[ \begin{aligned} \int \frac{1}{a+\sin\theta}d\theta &= \int \frac{1}{a+\cos\theta'}d\theta' \\ &= \frac{2}{a+1}\left[\tan\frac{\theta'}{2} - \sqrt{\frac{a-1}{a+1}}\arctan\frac{\tan\frac{\theta'}{2}}{\sqrt{\frac{a-1}{a+1}}} \right]+C \\ &= \frac{2}{a+1}\left[\tan\frac{2\theta-\pi}{4}- \sqrt{\frac{a-1}{a+1}}\arctan\frac{\tan\frac{2\theta-\pi}{4}}{\sqrt{\frac{a-1}{a+1}}} \right]+C \end{aligned} \]