«积分 $\int \frac{1}{\cos (x)+1} d x$ $\int \frac{1}{\cos (x)-1} d x$ and $\int \sqrt{1+\cos x} d x$»
by pluvet on Dec 14, 2019

$$ \begin{align} &\int \frac{1}{\cos (x)+1} d x\\ &=\int \frac{1{\color{red}(1-\cos x)}}{(1+\cos x){\color{red}(1-\cos x)}} d x\\ &=\int \frac{1-\cos x}{\sin ^{2} x} d x\\ &=\int \frac{1}{\sin ^{2} x}-\frac{\cos x}{\sin ^{2} x} d x\\ &=\int {\csc ^{2} x-\csc x \cdot \cot x d x} \\ &=\csc x-\cot x+C\\ &=tan(\frac{x}{2})+C \end{align} $$

$$ \begin{align} &\int \frac{1}{\cos (x)-1} d x\\ &=\int \frac{1{\color{red}(1+\cos x)}}{(1-\cos x){\color{red}(1+\cos x)}} d x\\ &=\int \frac{1+\cos x}{\sin ^{2} x} d x\\ &=\int \frac{1}{\sin ^{2} x}+\frac{\cos x}{\sin ^{2} x} d x\\ &=\int {\csc ^{2} x+\csc x \cdot \cot x d x} \\ &=-\cot x - \csc x+C\\ &=\frac{1}{tan(\frac{x}{2})}+C \end{align} $$

$$ \begin{aligned} & \int \sqrt{1+\cos x} d x \\ =& \int \sqrt{\frac{(1+\cos x)({\color{red}1-\cos x})}{{\color{red}1-\cos x}}} d x \\ =& \int \sqrt{\frac{\sin ^{2} x}{1-\cos x}} d x \\ =& \int \frac{\sin x d x}{\sqrt{1-\cos x}} \\ \text{Let}\ u=\cos x\\ \Rightarrow d u=-\sin x d x \\ =& \int-\frac{d u}{\sqrt{1-u}}\\ \text { Let } w=1-u \\ \Rightarrow d w=-d u \\ =& \int-\frac{d w}{\sqrt{w}}\\ =&\int w^{-\frac{1}{2}} d w\\ =&\frac{w^{\frac{1}{2}}}{\frac{1}{2}}+c\\ =&2 \sqrt{1-u}+c\\ =&2 \sqrt{1-\cos x}+c\\ \end{aligned} $$

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