# Thread: How to Integrate this?

1. ## How to Integrate this?

sec2xdx/cos^2[2x](sex2x-1)

Thanks!

2. Originally Posted by honestliar
sec2xdx/cos^2[2x](sex2x-1)

Thanks!
$Intg$( sec^3[2x]) $div$(sec2x - 1)

3. Hello, honestliar!

sa-ri-ga-ma gave you a big hint?
Was it enough?

Integrate: . $\int \frac{\sec(2x)\,dx}{\cos^2(2x)[\sec(2x)-1]}$
Since $\frac{1}{\cos(2x)} \:=\:\sec(2x)$

. . we have: . $\int\frac{\sec^3(2x)\,dx}{\sec(2x)-1} \;=\;\int\bigg[\sec^2(2x) + \sec(2x) + 1 + \frac{1}{\sec(2x) - 1}\bigg]\,dx$

The first three terms have standard integration formulas.

For the fraction: . $\frac{1}{\sec(2x) - 1} \;=\;\frac{1}{\frac{1}{\cos(2x)} - 1}$

Multiply by $\frac{\cos(2x)}{\cos(2x)}\!:\quad \frac{\cos(2x)}{\cos(2x)}\cdot\frac{1}{\frac{1}{\c os(2x)} - 1} \;=\;\frac{\cos(2x)}{1 - \cos(2x)}$

Multiply by $\frac{1+\cos(2x)}{1+\cos(2x)}\!:\quad \frac{1+\cos(2x)}{1+\cos(2x)}\cdot \frac{\cos(2x)}{1-\cos(2x)} \;=\;\frac{\cos(2x) - \cos^2(2x)}{1 - \cos^2(2x)}$

. . $=\;\;\frac{\cos(2x) - \cos^2(2x)}{\sin^2(2x)} \;\;=\;\;\frac{\cos(2x)}{\sin^2(2x)} - \frac{\cos^2(2x)}{\sin^2(2x)} \;\;=\;\;\frac{1}{\sin(2x)}\cdot\frac{\cos(2x)}{\s in(2x)} - \left(\frac{\cos(2x)}{\sin(2x)}\right)^2$

. . $= \;\;\csc(2x)\cot(2x) - \cot^2(2x) \;=\;\csc(2x)\cot(2x) - [\csc^2(2x) - 1]$

And we have: . $\int\bigg[\csc(2x)\cot(2x) - \csc^2(2x) + 1\bigg]\,dx$

. . which also have standard integration formulas.

4. Thank you very much, I got it

5. Or

$\int {\frac{{\sec 2x}}{{{{\cos }^2}2x\left( {\sec 2x - 1} \right)}}dx} .$

$\frac{{\sec 2x}}{{{{\cos }^2}2x\left( {\sec 2x - 1} \right)}} = \frac{1}{{\cos 2x}}\frac{1}{{{{\cos }^2}2x\left( {\frac{{1 - \cos 2x}}{{\cos 2x}}} \right)}} = \frac{1}{{{{\cos }^2}2x\left( {1 - \cos 2x} \right)}} =$

$= \frac{{1 - \cos 2x + \cos 2x}}{{{{\cos }^2}2x\left( {1 - \cos 2x} \right)}} = \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x\left( {1 - \cos 2x} \right)}} =$

$= \frac{1}{{{{\cos }^2}2x}} + \frac{{1 - \cos 2x + \cos 2x}}{{\cos 2x\left( {1 - \cos 2x} \right)}} = \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x}} + \frac{1}{{1 - \cos 2x}} =$

$= \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x}} + \frac{{1 + \cos 2x}}{{1 - {{\cos }^2}2x}} = \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x}} + \frac{{1 + \cos 2x}}{{{{\sin }^2}2x}} =$

$= \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x}} + \frac{1}
{{{{\sin }^2}2x}} + \frac{{\cos 2x}}{{{{\sin }^2}2x}}.$

$\int {\frac{{dx}}{{{{\cos }^2}2x}}} = \frac{1}{2}\int {\frac{{d\left( {2x} \right)}}{{{{\cos }^2}2x}}} = \frac{1}{2}\tan 2x + C.$

$\int {\frac{{dx}}{{\cos 2x}}} = \frac{1}{2}\int {\sec 2xd\left( {2x} \right)} = \frac{1}{2}\ln \left| {\tan 2x + \sec 2x} \right| + C.$

$\int {\frac{{dx}}{{{{\sin }^2}2x}}} = \frac{1}{2}\int {\frac{{d\left( {2x} \right)}}{{{{\sin }^2}2x}}} = - \frac{1}{2}\cot 2x + C.$

$\int {\frac{{\cos 2x}}{{{{\sin }^2}2x}}dx} = \left\{ \begin{gathered}\sin 2x = u, \hfill \\\cos 2xdx = \frac{{du}}{2} \hfill \\ \end{gathered} \right\} = \frac{1}{2}$ $\int {\frac{{du}}{{{u^2}}}} = - \frac{1}{{2u}} + C = - \frac{1}{{2\sin 2x}} + C = - \frac{1}{2}\csc 2x + C.$

Finally we have

$\int {\frac{{\sec 2x}}{{{{\cos }^2}2x\left( {\sec 2x - 1} \right)}}dx} = \frac{1}{2}\tan 2x - \frac{1}{2}\cot 2x - \frac{1}{2}\csc 2x + \frac{1}{2}\ln$ $\left| {\tan 2x + \sec 2x} \right| + C.$