Question

ok i have a question, im finding the surface areas of integrals. the questions is y=tanx, x=0, Pi/4, and on the x axis, i got the graph and i have the integral all setup i just need some help solving the integral, i started using u substitution, but there is were i get lost

Answer 1

I'm going to take a wild guess at what you mean...

You're given the curve \(y=\tan x\) over the interval \(\left[0,\dfrac{\pi}{4}\right]\), and you're considering the region bounded by this curve and the x-axis. This region is being revolved about the x-axis, and you want to find the surface area of this solid of revolution.

Is that correct?

If so, the integral setup I assume you have looks something like
\[A=2\pi\int_0^{\pi/4}\tan x\sqrt{1+\sec^4x}\,dx\]

Answer 2

yes, i just im confused on the solving part

Answer 3

At first I would have thought to do the following
\[\int_0^{\pi/4}\frac{\tan x\sec x}{\sec x}\sqrt{1+\sec^4x}\,dx=\int_1^{\sqrt2}\frac{\sqrt{1+y^4}}{y}\,dy\]
where \(y=\sec x\). This looks ... doable.

Answer 4

\[\int_1^{\sqrt2}\frac{y^3\sqrt{1+y^4}}{y^4}\,dy\]
Set \(u=1+y^4\), so \(du=4y^3\,dy\), and you have
\[\frac{1}{4}\int_2^5\frac{\sqrt{u}}{u-1}\,du\]

Answer 5

Set \(t=\sqrt u\), so that \(t^2=u\) and \(2t\,dt=du\).
\[\frac{1}{4}\int_2^5\frac{\sqrt u}{u-1}\,du=\frac{1}{2}\int_{\sqrt2}^{\sqrt5}\frac{t^2}{t^2-1}\,dt=\frac{1}{2}\int_{\sqrt2}^{\sqrt5}\left(1+\frac{1}{t^2-1}\right)\,dt\]
Partial fractions:
\[\frac{1}{t^2-1}=\frac{1}{(t-1)(t+1)}=\frac{1}{2}\left(\frac{1}{t-1}-\frac{1}{t+1}\right)\]
and so, finally,
\[A=2\pi\int_0^{\pi/4}\tan x\sqrt{1+\sec^4x}\,dx=\frac{\pi}{2}\int_\sqrt2^\sqrt5\left(2+\frac{1}{t-1}-\frac{1}{t+1}\right)\,dt\]