Question

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.

Answer 1

`\displaystyle\int\frac{√(x^2+1)}x\,\mathrm dx=\int(√(\tan^2B+1))/(\tan B)\sec^2B\,\mathrm dB`

`\tan^2B+1=\sec^2B`

`√(\sec^2B)=\sec B` (provided that
`\sec B>0`)

Then

`(\sec B)/(\tan B)\cdot\sec^2B=(\sec^2B)/(\tan B)\cdot\sec B`

(just moving around a factor of
`\sec B`)

`(\sec B)/(\tan B)\cdot\sec^2B=(\sec^2B)/(\tan^2B)\cdot\sec B\tan B`

(multiply by
`(\tan B)/(\tan B)`)