Getting rid of square root in this integration How do I get rid of this square root so that I can integrate it? 
$$S = 2 \pi a ^2 \int_0^{\sqrt{2}} t \sqrt{4t^2 + 1}\ dt$$
 A: Hint: Let $x=4t^2+1$. Then $dx=8t~dt$. Can you take it from here ? :-)
A: The substitution method is usually presented like this.
$$\require{cancel}\int_a^bu'(x)f(u(x))dx = \int_{u(a)}^{u(b)}f(u)du$$
I find that adding the intermediary steps makes the integration easier for me, so I multiply by
$$\frac{du/dx}{u'(x)}$$
Notice that the numerator and denominator are the same (i.e. the derivative of $u$ with respect to $x$), so the fraction evaluates to 1.
So lets multiply now
$$\int_a^bu'(x)f(u(x))dx=\int_{x=a}^{x=b}\frac{\cancel{u'(x)}f(u(x))\cancel{dx}\cdot\frac{du}{\cancel{dx}}}{\cancel{u'(x)}}= \int_{u(a)}^{u(b)}f(u)du$$
In addition. quite often I will express $du$ as (for example) $d(4x^2+1)$
$$\begin{array}{lll}
2\pi a^2\int_0^\sqrt{2}t\sqrt{4t^2+1}dt&=&2\pi a^2\int_0^\sqrt{2}t\sqrt{4t^2+1}dt\frac{d(4t^2+1)/dt}{d(4t^2+1)/dt}\\
&=&2\pi a^2\int_0^\sqrt{2}t\sqrt{4t^2+1}dt\frac{d(4t^2+1)/dt}{8t}\\
&=&2\pi a^2\int_0^\sqrt{2}t\sqrt{4t^2+1}\frac{d(4t^2+1)}{8t}\\
&=&2\pi a^2\int_0^\sqrt{2}\frac{t\sqrt{4t^2+1}}{8t}d(4t^2+1)\\
&=&2\pi a^2\int_0^\sqrt{2}\frac{\sqrt{4t^2+1}}{8}d(4t^2+1)\\
&=&\frac{1}{4}\pi a^2\int_0^\sqrt{2}\sqrt{4t^2+1}d(4t^2+1)\\
&=&\frac{1}{4}\pi a^2\int_1^9\sqrt{u}du\\
&=&\dots
\end{array}$$
Doing things this way keeps me focused on the problem without the need to go off to the side and calculate $du$.
