这是一份UCL伦敦大学 MATH6503 / MATHG653作业代写的成功案例
$$
T=\frac{1}{2 \omega} \int_{0}^{\theta_{1}} \frac{d \theta}{\sqrt{\left[\sin ^{2}(\alpha / 2)-\sin ^{2}(\theta / 2)\right]}}
$$
Now writing
$$
\sin (\theta / 2)=k \sin \phi
$$
where $k=\sin (\alpha / 2)$, then
$$
\omega T=\int_{0}^{\phi_{1}} \frac{d \phi}{\sqrt{\left(1-k^{2} \sin ^{2} \phi\right)}},
$$
in which $k \sin \phi_{1}=\sin \left(\theta_{1} / 2\right)$. The integral is referred to as an elliptic integral of the first kind and is usually denoted by
$$
F\left(k, \phi_{1}\right)=\int_{0}^{\phi_{1}} \frac{d \phi}{\sqrt{\left(1-k^{2} \sin ^{2} \phi\right)}},
$$
with $0 \leqslant k \leqslant 1$. Similarly, the elliptic integral of the second kind is defined by
$$
E\left(k, \phi_{1}\right)=\int_{0}^{\Phi_{1}} \sqrt[V]{\left(1-k^{2} \sin ^{2} \phi\right) d \phi}
$$
MATH6503 / MATHG653 COURSE NOTES ：
non-linear pendulum equation, by multiplying by $2 d y / d x$. Then
$$
2 \frac{d y}{d x} \frac{d^{2} y}{d x^{2}}=2\left(-a y-b y^{3}\right) \frac{d y}{d x}
$$
This equation may be integrated directly to give
$$
\left(\frac{d y}{d x}\right)^{2}=C-a y^{2}-\frac{1}{2} b y^{4}
$$
where $C$ is a constant of integration. Taking the square root of and separating the variables, we find
$$
\int \frac{d y}{\sqrt{\left(C-a y^{2}-\frac{1}{2} b y^{4}\right)}}=x+B
$$