How do you find the antiderivative of $\frac{cos (x)}{1 - cos (x)}$ $cos(x)/(1-cos(x))$ ?

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How do you find the antiderivative of $\frac{cos (x)}{1 - cos (x)}$ $cos(x)/(1-cos(x))$ ?

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Answer 1 · 2017-01-24T01:12:20

$- x - 2 cot (\frac{x}{2}) + C$ $-x-2cot(x/2)+C$

Explanation:

$I = \int \frac{cos (x)}{1 - cos (x)} d x$ $I=intcos(x)/(1-cos(x))dx$

Rewriting the integral in a simpler form:

$I = \int \frac{(cos (x) - 1) + 1}{1 - cos (x)} d x$ $I=int((cos(x)-1)+1)/(1-cos(x))dx$

$I = \int \frac{- (1 - cos (x))}{1 - cos (x)} d x + \int \frac{d x}{1 - cos (x)}$ $I=int(-(1-cos(x)))/(1-cos(x))dx+intdx/(1-cos(x))$

$I = - \int d x + \int \frac{d x}{1 - cos (x)}$ $I=-intdx+intdx/(1-cos(x))$

$I = - x + \int \frac{d x}{1 - cos (x)}$ $I=-x+intdx/(1-cos(x))$

For the remaining integral, we'll use the tangent half-angle substitution which uses $t = tan (\frac{x}{2})$ $t=tan(x/2)$ . The function $cos (x)$ $cos(x)$ can be expressed in terms of $tan (\frac{x}{2})$ $tan(x/2)$ as follows:

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That is, $cos (x) = \frac{1 - {tan}^{2} (\frac{x}{2})}{1 + {tan}^{2} (\frac{x}{2})} = \frac{1 - {tan}^{2} (\frac{x}{2})}{{sec}^{2} (\frac{x}{2})}$ $cos(x)=(1-tan^2(x/2))/(1+tan^2(x/2))=(1-tan^2(x/2))/sec^2(x/2)$ .

Also note that the substitution $t = tan (\frac{x}{2})$ $t=tan(x/2)$ implies $d t = \frac{1}{2} {sec}^{2} (\frac{x}{2}) d x$ $dt=1/2sec^2(x/2)dx$ .

Applying this to the integral gives:

$I = - x + \int \frac{d x}{(1 - \frac{1 - {tan}^{2} (\frac{x}{2})}{{sec}^{2} (\frac{x}{2})})}$ $I=-x+intdx/((1-(1-tan^2(x/2))/sec^2(x/2)))$

$I = - x + \int \frac{{sec}^{2} (\frac{x}{2}) d x}{{sec}^{2} (\frac{x}{2}) - (1 - {tan}^{2} (\frac{x}{2}))}$ $I=-x+int(sec^2(x/2)dx)/(sec^2(x/2)-(1-tan^2(x/2))$

Rewriting ${sec}^{2} (\frac{x}{2})$ $sec^2(x/2)$ as ${tan}^{2} (\frac{x}{2}) + 1$ $tan^2(x/2)+1$ :

$I = - x + \int \frac{{sec}^{2} (\frac{x}{2}) d x}{2 {tan}^{2} (\frac{x}{2})}$ $I=-x+int(sec^2(x/2)dx)/(2tan^2(x/2))$

$I = - x + 2 \int \frac{\frac{1}{2} {sec}^{2} (\frac{x}{2}) d x}{2 {tan}^{2} (\frac{x}{2})}$ $I=-x+2int(1/2sec^2(x/2)dx)/(2tan^2(x/2))$

Substituting:

$I = - x + 2 \int \frac{d t}{t^{2}}$ $I=-x+2intdt/t^2$

$I = - x - \frac{2}{t} + C$ $I=-x-2/t+C$

$I = - x - \frac{2}{tan (\frac{x}{2})} + C$ $I=-x-2/tan(x/2)+C$

$I = - x - 2 cot (\frac{x}{2}) + C$ $I=-x-2cot(x/2)+C$

Answer 2 · 2017-01-24T02:03:27

$- (x + cot (\frac{x}{2}) + C$ $-(x+cot(x/2) + C$

Explanation:

$\int \frac{cos x}{1 - cos x} d x = \int \frac{1 - 2 {sin}^{2} (\frac{x}{2})}{2 {sin}^{2} (\frac{x}{2})} d x$ $int cos x/(1-cos x) dx=int (1-2sin^2(x/2))/(2sin^2(x/2)) dx$

$= \int \frac{1}{2} {csc}^{2} (\frac{x}{2}) d x - \int d x$ $=int1/2 csc^2(x/2) dx-int dx$

$= \int d (- cot (\frac{x}{2}) - x$ $=intd(-cot(x/2) -x$

$= - (x + cot (\frac{x}{2})) + C$ $=-(x+cot(x/2)) + C$ .

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