Question

int (2x-sin 2x)/(1-cos 2x) dx

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int \frac{2x - sin2x}{1 - cos2x} \: dx$

Let we assume that,

$\rm :\longmapsto\:I = \displaystyle\int \frac{2x - sin2x}{1 - cos2x} \: dx$

Now, it can be rewritten as

$\rm :\longmapsto\:I = \displaystyle\int \frac{2x}{1 - cos2x} \: dx - \displaystyle\int \frac{sin2x}{1 - cos2x} \: dx$

Let suppose that,

$\rm :\longmapsto\:I = I_1 - I_2 - - - (1)$

where,

$\rm :\longmapsto\:I_1 = \displaystyle\int \frac{2x}{1 - cos2x} \: dx$

and

$\rm :\longmapsto\:I_2 = \displaystyle\int \frac{sin2x}{1 - cos2x} \: dx$

Now, Consider

$\rm :\longmapsto\:I_2 = \displaystyle\int \frac{sin2x}{1 - cos2x} \: dx$

$\rm :\longmapsto\:I_2 = \displaystyle\int \frac{2sinxcosx}{2 {sin}^{2}x } \: dx$

$\rm :\longmapsto\:I_2 = \displaystyle\int \frac{cosx}{sinx} \: dx$

$\rm :\longmapsto\:I_2 = \displaystyle\int cotx\: dx$

$\bf\implies \:I_2 = logsinx + c_1 - - - - (2)$

Now, Consider

$\rm :\longmapsto\:I_1 = \displaystyle\int \frac{2x}{1 - cos2x} \: dx$

$\rm :\longmapsto\:I_1 = \displaystyle\int \frac{2x}{ {2sin}^{2} x} \: dx$

$\rm :\longmapsto\:I_1 = \displaystyle\int \frac{x}{ {sin}^{2} x} \: dx$

$\rm :\longmapsto\:I_1 = \displaystyle\int x {cosec}^{2}x \: dx$

We know,

Integrating by parts,

$\red{ \boxed{ \sf{ \:\displaystyle\int uvdx = u\displaystyle\int vdx - \displaystyle\int \bigg[\dfrac{d}{dx}u\displaystyle\int vdx\bigg]dx}}}$

So, here

$\red{\rm :\longmapsto\:u = x}$

and

$\red{\rm :\longmapsto\:v = cosecx}$

So, on substituting the values, we get

$\rm \:  =  \: x\displaystyle\int {cosec}^{2}xdx - \displaystyle\int \bigg[\dfrac{d}{dx}x\displaystyle\int {cosec}^{2}xdx\bigg]dx$

$\rm \:  =  \: - x \: cotx \: + \displaystyle\int cotx \: dx$

$\rm \:  =  \: - x \: cotx \: + logsinx \: + \: c_2$

So, on substituting the values in equation (1), we get

$\rm :\longmapsto\:I = - xcotx + logsinx + c_1 - logsinx - c_2$

$\bf\implies \:I = - x \: cotx \: + \: c$

Additional Information :

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$