Find the surface area of revolution when the line $y$ $=$ $cos((1/2)x)$ is rotated about the x-axis from $0\le x\le \pi$.
So I got the integral to be:
$2\pi \int _0^{\pi }\:cos\left(\frac{1}{2}x\right)\sqrt{1+\frac{1}{4}sin^2\left(\frac{1}{2}x\right)}dx$
Using $u\:=\:sin\left(\frac{1}{2}x\right)$ and $du=\frac{1}{2}cos\left(\frac{1}{2}x\right)dx$, I got:
$2\pi \int _0^1\:\sqrt{1+\frac{1}{4}u^2}2du=2\pi \int _0^1\:\sqrt{4+u^2}du$
The using the trig sub of $u\:=\:2tan\theta $ I got:
$8\pi \int _0^1\:sec^3\theta$
Which became:
$4\pi \left(ln\left|tan\theta +sec\theta \right|+sec\theta tan\theta \right)$ for $0\le x\le \pi$
However, after this I got stuck. I am not even sure if the last couple of steps I have done above are correct. I am also a little confused because of the amount of substitutions. I am not sure how to plug in the appropriate values again.
Any help?
$u=2 \tan \theta$,
when $u=0, \theta = 0$.
when $u = 1$, $\theta = \tan^{-1} \frac12$
You just have to substitute these corresponding values to evaluate the value.