I am fairly new to calculus and self learning integration from home has been challenging so I'm sorry if I make any mistakes. I want to work out the arc length of:
$y = \sqrt{7.2 (x-\frac {1}{7}}) - 2.023, [0.213, 0.1.27]$.
I have used the definition of a definite integral and got
$\int_{0.075}^{0.58} \sqrt{1+\left(\frac{3.54}{\sqrt{7x-1}}\right)²} dx$ =$\int_{0.075}^{0.58} \sqrt{1+\frac{12.5316}{7x-1}} dx$
So far which I think is correct. How would I proceed from here? Would I use u-substitution and then a trigonometric substitution to eliminate the exponent? Any help is appreciated.
First hint: let $C = 12.5316$, and never write any more decimals in the rest of your work. You can plug in $C$ at the end. Similarly, I'll use $a$ and $b$ for the limits of integration. So you have \begin{align} I &= \int_a^b \sqrt{1 + \frac{C}{7x-1}}~dx \end{align} First step is common denominator: \begin{align} I &= \int_a^b \sqrt{\frac{7x -1 + C}{7x-1}}~dx \end{align} Now let $u = 7x + 1, du = 7 dx$ to get \begin{align} I &= \int_a^b \sqrt{\frac{7x -1 + C}{7x-1}}~dx\\ &= \int_{x=a}^{x=b} \sqrt{\frac{u + C}{u}}~\frac{1}{7}~du\\ \text{so that}\\ 7I &= \int_{x=a}^{x=b} \sqrt{\frac{u + C}{u}}~du\\ \end{align}
At this point...you'd really like to get rid of that $C$, so let's let $$ u = C\tan^2 v \\ du = 2C \tan v \sec^2 v ~ dv $$ so that \begin{align} 7I &= \int_{x=a}^{x=b} \sqrt{\frac{u + C}{u}}~du\\ &= \int_{x=a}^{x=b} \sqrt{\frac{C(\tan^2 v + 1)}{C \tan^2 v}}~2C\tan v \sec^2 v ~ dv\\ &= 2C\int_{x=a}^{x=b} \sqrt{\frac{\sec^2 v}{\tan^2 v}}\tan v \sec^2 v ~ dv; & \text{write out in sines/cosines...}\\ &= 2C\int_{x=a}^{x=b} \sqrt{\frac{\frac{1}{\cos^2 v}}{\frac{\sin^2 v}{\cos^2 v}}}\frac{\sin v}{\cos^3 v}~ dv; \\ &= 2C\int_{x=a}^{x=b} \sqrt{\frac{1}{\sin^2 v}}\frac{\sin v}{\cos^3 v}~ dv \\ &= 2C\int_{x=a}^{x=b} \frac{1}{\sin v}\frac{\sin v}{\cos^3 v}~ dv \\ &= 2C\int_{x=a}^{x=b} \frac{1}{\cos^3 v}~ dv \\ &= 2C\int_{x=a}^{x=b} \frac{\cos v}{\cos^4 v}~ dv \\ &= 2C\int_{x=a}^{x=b} \frac{\cos v}{(1-\sin^2(v))^2}~ dv; & \text{subst $t = \sin v$} \\ &= 2C\int_{x=a}^{x=b} \frac{1}{(1-t^2)^2}~ dt; \\ \end{align} which is now an ordinary partial-fractions integral; when you're done, you'll have to backsubstitute for $t$, and then $v$, and then $u$ to finally get an expression in $x$ into which you can plug the limits $a$ and $b$.
This does feel like a long way around, and there's probably a far simpler approach...but this is the one that jumped out at me.