# Length of a function using integrals

• February 24th 2006, 04:57 PM
alysha27
Length of a function using integrals
Find the length of the cardioid r=1+sinO (sorry I wish i knew how to use all the symbols)

I know that

L= integral of [r^2 +(dr/dO)^2]^1/2 dO (from 0 to 2pi)

= integral of [1 + 2sinO + sin^2(O) +cos^2(O)]^1/2 dO...

=integral of (2+2sinO)^1/2 dO from 0 to 2pi,

... then I get stuck on simplifying do I can something I can integrate. I know to multiplying and dividing by the integrand (2-2sinO)^1/2. Can't seem to remember how to do this from high school. Any help would be great. Thanks.
• February 25th 2006, 10:56 AM
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:

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

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

$
\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:

$
\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.

$\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:

$
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
$
• February 25th 2006, 03:01 PM
topsquark
Quote:

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:

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

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

$
\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:

$
\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.

$\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:

$
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
$

Hey that was cool! :cool: I've never seen that trick before.

-Dan