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I have been asked to find the surface area formed when $y=\cos(x/2)$ is rotated around the $x-$axis from $x=0$ to $\pi$. I understand how to set up the integral, but I am really struggling solving it. Here is how far I have been able to go so far:

$$2\pi \int _0^{\pi }\cos\left(\frac{x}{2}\right)\sqrt{1+\left(-\frac{1}{2}\sin\left(\frac{x}{2}\right)\right)^2}$$

$$2\pi \int _0^{\pi }\cos\left(\frac{x}{2}\right)\sqrt{1+\frac{1}{4}\sin^2\left(\frac{x}{2}\right)}$$

$$\pi \int _0^{\pi }\cos\left(\frac{x}{2}\right)\sqrt{4+\sin^2\left(\frac{x}{2}\right)}$$

From here I figured that I needed to do some sort of trig sub, but none of the things I have tried have worked.

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Here's an easy way to solve this, pretty algorithmic - not the fastest by far, but easy to follow and carry out in general $$\pi \int _0^{\pi }\cos\left(\frac{x}{2}\right)\sqrt{4+\sin^2\left(\frac{x}{2}\right)}\,dx$$ Let $\frac{x}{2} = u \implies dx = 2du$ $$2\pi \int _0^{\frac{\pi}{2} }\cos\left(u\right)\sqrt{4+\sin^2\left(u\right)}\,du$$ Let $\sin u = v \implies dv = \cos (u) \,du$ $$2\pi \int_0^1 \sqrt{4+v^2}\,dv$$ The subsitution $v = 2p$ yields $$8\pi \int_0^\frac{1}{2} \sqrt{1+p^2}\,dp$$ The trig subs from here shouldn't be bad!

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$$\int_0^{\pi} \cos \left(\frac x2 \right) \sqrt{4+\sin^2 \left(\frac x2 \right)} dx$$

Let $\sin \left( \frac x2 \right) = 2\tan (\theta)$

$\frac 12 \cos \left( \frac x2 \right) dx = 2\sec^2 (\theta) d\theta$

Then the integral becomes

$$\int_0^{\arctan \left(\frac 12 \right)} 8\sec^3 \theta d\theta$$

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