Integrating $\int _0^1\sqrt{9\sin^2(x)+9\cos^2(x)+16}\ \mathrm dx$ What I'm trying to do is find the arc length of $r(x) = \langle 3\cos(x),4x,3\sin(x)\rangle$ when $0\le x\le 1$.
After I derived the vector and found the magnitude of that derived vector, I came up with $$\int_0^1\sqrt{9\sin^2(x)+9\cos^2(x)+16}\ \mathrm dx$$
as the solution. Now I have to integrate it and I'm not so sure where to begin because I haven't seen a problem like this before.
Hint
$\sin^2 x+\cos^2 x = 1$ due to Pythagoras, so the question is
$$\int_0^1 \sqrt{25}\ \mathrm dx = ?$$



Since there has been a discussion if the integral was actually correct (wich it was), here is the long answer:
$$r(x) = \langle 3\cos x, 4x, 3\sin x\rangle\\
\nabla r(x) = \langle -3\sin x, 4, 3\cos x\rangle$$
Now the arc length is defined by
$$L_a^b (r) := \int_a^b \| \nabla r(x) \|_2\ \mathrm dx$$
Note that $\|x\|_2 = \sqrt{x_1^2 + x_2^2 + \ldots}$ so
$$\|\nabla r(x)\|_2 = \sqrt{(-3\sin x)^2 + 4^2 + (3\cos x)^2} = \sqrt{9 (\sin^2 x + \cos^2 x) + 16} = \sqrt{25} = 5$$
Where we used Pythagoras ($\sin^2 x +\cos^2 x = 1$ for all $x$). Thus
$$L_0^1 (r) = \int_0^1 5\ \mathrm dx = 5$$
$$
\int_{0}^{1}\sqrt{9\sin(x)^2+16\cos(x)^2+16}\,\, dx \\
\int_{0}^{1}\sqrt{9(\sin(x)^2+\cos(x)^2)+16}\,\,dx  \\
\left[\sin^2(x)+\cos^2(x) = 1\right] \\
\int_{0}^{1}\sqrt{9+16}\,\,dx = \int_{0}^{1}\sqrt{25}\,\,dx = \int_{0}^{1}5\,\,dx = 5*(1-0) = 5
$$