How do I get rid of this square root so that I can integrate it?
$$S = 2 \pi a ^2 \int_0^{\sqrt{2}} t \sqrt{4t^2 + 1}\ dt$$
$\endgroup$ 12 Answers
$\begingroup$Hint: Let $x=4t^2+1$. Then $dx=8t~dt$. Can you take it from here ? :-)
$\endgroup$ 1 $\begingroup$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 $du$.
$\endgroup$
