# Thread: Integration Problem

1. ## Integration Problem

Problem: Express the arc length of the following curves in terms of the integral E(k) = $\int \sqrt{1 - k^2 sin^2 (t)} dt$ (0 < k < 1) and the integral runs from 0 to $\pi/2$, for suitable values of k. (E(k) is one of the standard elliptic integrals, so called because of their connection with the arc length of an ellipse.)
(a) An ellipse with semimajor axis a and semiminor axis b.
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I've solved for the appropriate function and now I'm just left with the integration. I have the following (L(C) denotes the arc length):
L(C) = 4 $\int \sqrt{a^2 sin^2 (t) + b^2 cos^2 (t)} dt$

L(C) = 4 $\int \sqrt{\frac{a^2 (1 - cos(2t)) + b^2 (1 + cos(2t))}{2}} dt$
$\vdots$
A handful of steps later, I get
L(C) = 4 $\int \sqrt{b^2 - (b^2 - a^2)sin^2 (t)} dt$

Sad to say but I'm actually stuck here. Can anyone shed some light on how to finish this? Thank you.

2. ## Re: Integration Problem

Originally Posted by MissMousey
Problem: Express the arc length of the following curves in terms of the integral E(k) = $\int \sqrt{1 - k^2 sin^2 (t)} dt$ (0 < k < 1) and the integral runs from 0 to $\pi/2$, for suitable values of k. (E(k) is one of the standard elliptic integrals, so called because of their connection with the arc length of an ellipse.)
(a) An ellipse with semimajor axis a and semiminor axis b.
-----------------------------------------------------------------------

I've solved for the appropriate function and now I'm just left with the integration. I have the following (L(C) denotes the arc length):
L(C) = 4 $\int \sqrt{a^2 sin^2 (t) + b^2 cos^2 (t)} dt$

L(C) = 4 $\int \sqrt{\frac{a^2 (1 - cos(2t)) + b^2 (1 + cos(2t))}{2}} dt$
$\vdots$

A handful of steps later, I get

L(C) = 4 $\int \sqrt{b^2 - (b^2 - a^2)sin^2 (t)} dt$

Sad to say but I'm actually stuck here. Can anyone shed some light on how to finish this? Thank you.

Take b^2 out of the square root. Note: $k^2 = \frac{b^2 - a^2}{b^2}$