Help with cal 2 exercise involving integrals?
I have to find the arc length of y = lnx given that 1 ≤ x ≤ sqrt (3)
Here is my procedure:

I found the derivative of y = lnx
And got, y' = 1 / (x)
1 + (1/(x))^2 = 1 + (1/(x^2)) = (x^2+1)/(x^2)
integral from 1 to sqrt 3 of sqrt((x^2+1/(x^2)) dx = sqrt ( x ^ 2 + 1 ) / ( x ) dx
X = tan b
Dx = (sec^2)BdB
Sqrt(x^2+1) = Sqrt ( (tan^2)B + 1) = Sqrt ( (Sec^2)B) = SecB

Integral = ((sec)/(tanB)) times (sec^2)BdB

And here comes the part I don't understand, in my exercise it becomes

Integral = ((sec^2)B)/((tan^2)B)) times secBtanBdB

Why did this happen, please help me figure out how.

Question

contestada

Respuesta :

LammettHash LammettHash · Answer 1 · 2019-04-02T23:06:54+02:00

[tex]\displaystyle\int\frac{\sqrt{x^2+1}}x\,\mathrm dx=\int\frac{\sqrt{\tan^2B+1}}{\tan B}\sec^2B\,\mathrm dB[/tex]

[tex]\tan^2B+1=\sec^2B[/tex]

[tex]\sqrt{\sec^2B}=\sec B[/tex] (provided that [tex]\sec B>0[/tex])

Then

[tex]\dfrac{\sec B}{\tan B}\cdot\sec^2B=\dfrac{\sec^2B}{\tan B}\cdot\sec B[/tex]

(just moving around a factor of [tex]\sec B[/tex])

[tex]\dfrac{\sec B}{\tan B}\cdot\sec^2B=\dfrac{\sec^2B}{\tan^2B}\cdot\sec B\tan B[/tex]

(multiply by [tex]\dfrac{\tan B}{\tan B}[/tex])