Integration 2

\hspace{-16}$For $\bf{(3)::}$ Given $\bf{\int \tan(x-k)\cdot \tan(x+k)\cdot \tan(2x)dx}$\\\\\\ Using The formula $\bf{\bullet \;\; \tan\left\{(x+k)+(x-k)\right\}=\frac{\tan(x+k)+\tan(x-k)}{1-\tan(x+k)\cdot \tan(x-k)}}$\\\\\\ So $\bf{\tan(2x)=\frac{\tan(x+k)+\tan(x-k)}{1-\tan(x+k)\cdot \tan(x-k)}}$\\\\\\ So $\bf{\tan(x-k)\cdot \tan(x+k)\cdot \tan(2x) = \tan(2x)-\tan(x+k)-\tan(x-k)}$\\\\\\ So $\bf{\int \tan(x-k)\cdot \tan(x+k)\cdot \tan(2x)dx = \int \tan(2x)dx-\int \tan(x+k)dx-\int \tan(x-k)dx}$\\\\\\ So $\bf{\int \tan(x-k)\cdot \tan(x+k)\cdot \tan(2x)dx =+\frac{1}{2}\cdot \ln\left|\sec(2x)\right|-\ln\left|\sec(x+k)\right|-\ln\left|\sec(x-k)\right|+\mathbb{C}}$

\hspace{-16}$For $\bf{(2)::}$Given $\bf{\int\frac{3\cot 3x -\cot x}{\tan x-3\tan 3x}dx = \int\frac{3\tan x-\tan 3x}{\tan x-3\tan 3x}\cdot \frac{1}{\tan x\cdot \tan 3x}dx}$\\\\\\ Using $\bf{\bullet \; \tan 3x = \frac{3\tan x-\tan^3 x}{1-3\tan^2 x}}$\\\\\\ $\bf{ = \int\frac{1-3\tan^2 x}{3-\tan^2 x}dx = \int\frac{3\left(3-\tan^2 x\right)-8}{3-\tan^2 x}dx}$\\\\\\ $\bf{ = 3\int 1 dx -8\int\frac{1}{3-\tan^2x}dx}$\\\\\\ Now Let $\bf{\tan x= t}$ and $\bf{\sec^2 xdx = dt\Rightarrow dx = \frac{1}{1+\tan^2 x}dt = \frac{1}{1+t^2}dt}$\\\\\\ $\bf{= 3x-8\int\frac{1}{\left(1+t^2\right)\cdot \left(3-t^2\right)}dt = 3x+8\int\frac{1}{\left(1+t^2\right)\cdot \left(t^2-3\right)}dt}$\\\\\\

\hspace{-16}\bf{= 3x+8\cdot \frac{1}{4}\int\frac{1}{t^2-3}dt-\frac{8}{4}\int\frac{1}{1+t^2}dt}$\\\\\\ $\bf{ = 3x+2\cdot \ln \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|-2\cdot \tan^{-1}\left(t\right)+\mathbb{C}}$\\\\\\ So $\bf{\int\frac{3\cot 3x -\cot x}{\tan x-3\tan 3x}dx = 3x+2\cdot \ln \left|\frac{\tan x-\sqrt{3}}{\tan x+\sqrt{3}}\right|-2\cdot \tan^{-1}\left(\tan x\right)+\mathbb{C}}$

\hspace{-16}$For $\bf{(1)::}$Given $\bf{\int\frac{1}{\cos x+\csc x}dx = \int \frac{\sin x}{\sin x\cdot \cos x+1}}$\\\\\\ Now Multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\bf{2\;,}$ We get\\\\\\ So Integral is $\bf{=\int\frac{(\sin x+\cos x)+(\sin x-\cos x)}{\sin 2x+2}dx}$\\\\\\ So Integral is $\bf{\int\frac{\left(\sin x+\cos x\right)}{\sin 2x+2}dx+\int\frac{\left(\sin x-\cos x\right)}{\sin 2x+2}dx}$\\\\\\ Now Let $\bf{I = \int\frac{\left(\sin x+\cos x\right)}{\sin 2x+2}dx = \int\frac{\left(\sin x+\cos x\right)}{\left(\sqrt{3}\right)^2-\left(\sin x- \cos x\right)^2}dx}$\\\\\\ Now Let $\bf{\left(\sin x- \cos x\right)=t\;,}$ Then $\bf{\left(\sin x +\cos x\right)dx = dt}$\\\\\\ So $\bf{I = \int\frac{1}{\left(\sqrt{3}\right)^2-t^2}dt = \frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}+\left(\sin x- \cos x\right)}{\sqrt{3}-\left(\sin x- \cos x\right)}\right|+\mathbb{C_{1}}}$\\\\\\

\hspace{-16}$Similarly Let $\bf{J = \int\frac{\left(\sin x-\cos x\right)}{\sin 2x+2}dx = \int \frac{\left(\sin x-\cos x\right)}{\left(\sin x+\cos x\right)^2+1^2}dx}$\\\\\\ Now Let $\bf{\left(\sin x+\cos x\right)=u\;,}$ Then $\bf{\left(\sin x-\cos x\right)dx = -du}$\\\\\\ So Integral $\bf{J = -\int\frac{1}{1+u^2}du = -\tan^{-1}\left(\sin x+\cos x\right)+\mathbb{C_{2}}}$\\\\\\ So $\bf{\int \frac{1}{\cos x+\csc x}dx = \frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}+\left(\sin x- \cos x\right)}{\sqrt{3}-\left(\sin x- \cos x\right)}\right|-\tan^{-1}\left(\sin x+\cos x\right)+\mathbb{C}}$