Surface Area (Integration) Find the surface area of the object by rotating $y = 4+3x^2$ about the $y$-axis, where $1 \leq x \leq 2$.
I've been using the formula:
Surface area: 
Definite integral of $$2\pi y\Bigg(1+\Big(\dfrac{dx}{dy}\Big)^2\Bigg)dy$$
Limits are supposed to be in the form of $y$ for the above question.
I've been trying to do this question for the last 3 hours and I get stuck at $$\int\dfrac{\cos(x)}{\sin^5(x)}dx$$
Can anyone help me how to solve the above form?
Thanks in advance. 

Since $x=\sqrt{\frac{y-4}{3}}, \;\;\frac{dx}{dy}=\frac{1}{2}\left(\frac{y-4}{3}\right)^{-\frac{1}{2}}\cdot\frac{1}{3}\text{ and }\left(\frac{dx}{dy}\right)^2=\frac{1}{36}\frac{1}{\frac{y-4}{3}}=\frac{1}{12(y-4)}$.
Then $S=\displaystyle\int_7^{16} 2\pi\sqrt{\frac{y-4}{3}}\sqrt{1+\frac{1}{12(y-4)}}dy=2\pi\int_7^{16}\sqrt{\frac{y-4}{3}}\sqrt{\frac{12y-47}{12(y-4)}}dy$, 
so $\;\;S=\displaystyle\frac{\pi}{3}\int_7^{16}\sqrt{12y-47}dy$.  $\;\;\;$ (Now let $u=12y-47$.)

Alternatively, $S=\displaystyle\int_1^2 2\pi x\sqrt{1+36x^2}dx$, 
so letting $u=1+36x^2$ and $du=72xdx$ gives
$\;\;\displaystyle S=\frac{\pi}{36}\int_{37}^{145}\sqrt{u}du$.
