Another integral

[wondering]

\(\displaystyle \int\frac{2}{3(secx-1)}dx\)

\(\displaystyle \frac{2}{3}\int\frac{1}{secx-1}*\frac{secx+1}{secx+1}dx\)

\(\displaystyle \frac{2}{3}\int\frac{secx+1}{sec^2x-1}dx\)

\(\displaystyle \frac{2}{3}\int\frac{secx+1}{tan^2x}dx\)

Not sure if I'm on the right path or not. I don't see where to go from here. TIA

 

[DrPhil]

\(\displaystyle \int\frac{2}{3(secx-1)}dx\)

\(\displaystyle \frac{2}{3}\int\frac{1}{secx-1}*\frac{secx+1}{secx+1}dx\)

\(\displaystyle \frac{2}{3}\int\frac{secx+1}{sec^2x-1}dx\)

\(\displaystyle \frac{2}{3}\int\frac{secx+1}{tan^2x}dx\)

Not sure if I'm on the right path or not. I don't see where to go from here. TIA

I don't know if this is the "best" path, but it looks promising to me. The next step is to separate the numerator into two integrals.

\(\displaystyle \displaystyle \frac{2}{3}\int \dfrac{\sec{x}}{\tan^2{x}}dx + \frac{2}{3}\int \dfrac{1}{\tan^2{x}}dx\)

The first integrand becomes \(\displaystyle \cos(x)/\sin^2(x)\), which is straightforward to integrate because the numerator is the derivative of the denominator. The second is \(\displaystyle \cos^2(x)/\sin^2(x)\), which is not trivial - but IS in the table of integrals that I use. If you don't have such a table, they can (probably) be found by integration by parts.

[Thanks to srmichael for pointing out error in 1st integrand]

 

[Deleted member 4993]

\(\displaystyle \int\frac{2}{3(secx-1)}dx\)

\(\displaystyle \frac{2}{3}\int\frac{1}{secx-1}*\frac{secx+1}{secx+1}dx\)

\(\displaystyle \frac{2}{3}\int\frac{secx+1}{sec^2x-1}dx\)

\(\displaystyle \frac{2}{3}\int\frac{secx+1}{tan^2x}dx\)

Not sure if I'm on the right path or not. I don't see where to go from here. TIA

Anotherway:

\(\displaystyle \displaystyle \int\frac{2}{3(secx-1)}dx\)

= \(\displaystyle \displaystyle\frac{2}{3} \int\frac{cos(x)}{1-cos(x)}dx\)

= \(\displaystyle \displaystyle\frac{2}{3} \int\frac{1-2sin^2(\frac{x}{2})}{2sin^2(\frac{x}{2})}dx\)

= \(\displaystyle \displaystyle\frac{2}{3} \int \left [ \frac{1}{2}cosec^2(\frac{x}{2}) - 1 \right ]dx\)

= \(\displaystyle \displaystyle\frac{2}{3}\left [-cot(\frac{x}{2}) - x \right ] + C \)

 

[wondering]

Thank you both for the suggestions. I will have to play around with the problem some more tonight to see if I could solve it on my own. Thanks again.

 

[soroban]

Hello, wondering!

And yet another way . . .

\(\displaystyle \displaystyle\int\frac{2}{3(\sec x-1)}\,dx \;=\;\frac{2}{3}\int\frac{dx}{\sec x-1} \)

We have: .\(\displaystyle \displaystyle \frac{1}{\sec x - 1} \;=\; \frac{1}{\frac{1}{\cos x} - 1} \;=\;\frac{\cos x}{1-\cos x} \)

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

. . \(\displaystyle \displaystyle =\;\frac{\cos x + \cos^2\!x}{\sin^2\!x} \;=\; \frac{\cos x}{\sin^2\!x} + \frac{\cos^2\!x}{\sin^2\!x} \;=\;\frac{1}{\sin x}\!\cdot\!\frac{\cos x}{\sin x} + \left(\frac{\cos x}{\sin x}\right)^2\)

. . \(\displaystyle =\;\csc x\cot x + \cot^2\!x \;=\;\csc x\cot x + \csc^2\!x -1\)


We have: .\(\displaystyle \displaystyle\frac{2}{3}\int\left(\csc x\cot x + \csc^2\!x - 1\right)\,dx\)

. . . . . . . . \(\displaystyle \displaystyle=\;\frac{2}{3}\big(-\csc x - \cot x - x\big) + C \)

. . . . . . . . \(\displaystyle =\;-\dfrac{2}{3}\big(\csc x + \cot x + x\big) + C\)

 

[Deleted member 4993]

Corollary from above:

cot(x/2) = csc(x) + cot(x)

did not see that before!!!