Originally Posted by

**TD!** You could do it using substitution I assume (haven't checked), but here's an approach using some trigonometry tricks.

I'm shifting t to pi/2-t. Of course, the boundaries shift too:

$\displaystyle \int\limits_0^{2\pi } {\sqrt {2 + 2\sin t} } dt = \int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\sin \left( {\frac{\pi }{2} - t} \right)} } dt$

Now we can use $\displaystyle \sin \left( \pi/2 - \alpha } \right) = \cos \alpha $ and then convert to the half angle using $\displaystyle \cos \left( {2\alpha } \right) = 2\cos ^2 \alpha - 1$

$\displaystyle

\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\cos t} } dt = \int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\left( {2\cos ^2 \left( {\frac{t}{2}} \right) - 1} \right)} } dt

$

Now we solved two problems: we introduced a square and we can get rid off the constant 2:

$\displaystyle

\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\left( {2\cos ^2 \left( {\frac{t}{2}} \right) - 1} \right)} } dt\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {4\cos ^2 \left( {\frac{t}{2}} \right)} } dt

$

Now we can lose the square root, but watch out that it's the absolute value which comes out of the square root. The problem now is that cos(t/2) becomes negative in our interval for t. But the area under cos(t/2) is the same on the interval [-pi/2,3pi/2] as on [-pi,pi] and on this last one, cos(t/2) is positive so we solved our problem of the absolute value without needing to integrate it.

$\displaystyle \int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {4\cos ^2 \left( {\frac{t}{2}} \right)} } dt = 2\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left| {\cos \left( {\frac{t}{2}} \right)} \right|} dt = 2\int\limits_{ - \pi }^\pi {\cos \left( {\frac{t}{2}} \right)} dt

$

And now it's easy:

$\displaystyle

2\int\limits_{ - \pi }^\pi {\cos \left( {\frac{t}{2}} \right)} dt = \left[ {4\sin \left( {\frac{t}{2}} \right)} \right]_{ - \pi }^\pi = 4 + 4 = 8

$