Integration Question (1 Viewer)

pikachu975 · Feb 13, 2017

Indefinite integral of (5x+1)dx / sqrt(4-x^2)

By splitting the numerator method please!

This is the question, but if anyone wants to see it is also Terry Lee 4.3 Q1c

InteGrand · Feb 13, 2017

pikachu975 said:
Indefinite integral of (5x+1)dx / sqrt(4-x^2)

By splitting the numerator method please!

This is the question, but if anyone wants to see it is also Terry Lee 4.3 Q1c

$\noindent By \textbf{inspection}, the answer is $-5\sqrt{4-x^{2}} + \sin^{-1}\left(\frac{x}{2}\right)+c$.$

$\noindent $\Big{(}$Split up the numerator and you're left with two standard integrals, i.e. $\int \frac{5x}{\sqrt{4-x^{2}}}\,\mathrm{d}x + \int \frac{1}{\sqrt{4-x^{2}}}\,\mathrm{d}x$. If you can't see how to do the first one by inspection, just substitute $u = 4-x^{2}$.$\Big{)}$$

pikachu975 · Feb 14, 2017

InteGrand said:
$\noindent By \textbf{inspection}, the answer is $-\frac{5}{2}\sqrt{4-x^{2}} + \sin^{-1}\left(\frac{x}{2}\right)+c$.$

$\noindent (Split up the numerator and you're left with two standard integrals. If you can't see how to do the first one by inspection, just substitute $u = 4-x^{2}$.)$

Yeah the first one was the one I couldn't do, but I guess it's just using the form that dsqrt(2x)/dx = -2/sqrt(x) right?

InteGrand · Feb 14, 2017

pikachu975 said:
Yeah the first one was the one I couldn't do, but I guess it's just using the form that dsqrt(2x)/dx = -2/sqrt(x) right?

$\noindent If we substitute $u = 4-x^{2}$, then $\mathrm{d}u = -2x\,\mathrm{d}x \Rightarrow 5x \,\mathrm{d}x = -\frac{5}{2}\,\mathrm{d}u$. Hence$

$\begin{align*}\int \frac{5x}{4-x^{2}}\,\mathrm{d}x &= -\frac{5}{2}\int \frac{\mathrm{d}u}{\sqrt{u}} \\ &= -\frac{5}{2}\times 2\sqrt{u}\\ &= -5u \\ &= -5\sqrt{4-x^{2}}.\end{align*}$

pikachu975 · Feb 14, 2017

InteGrand said:
$\noindent If we substitute $u = 4-x^{2}$, then $\mathrm{d}u = -2x\,\mathrm{d}x \Rightarrow 5x \,\mathrm{d}x = -\frac{5}{2}\,\mathrm{d}u$. Hence$

$\begin{align*}\int \frac{5x}{4-x^{2}}\,\mathrm{d}x &= -\frac{5}{2}\int \frac{\mathrm{d}u}{\sqrt{u}} \\ &= -\frac{5}{2}\times 2\sqrt{u}\\ &= -5u \\ &= -5\sqrt{4-x^{2}}.\end{align*}$

Thanks for the help.

Can they still ask you to integrate like 1/sqrt(x^2 - 4) which is part of the table of standard integrals (not given anymore)?

InteGrand · Feb 14, 2017

pikachu975 said:
Thanks for the help.

Can they still ask you to integrate like 1/sqrt(x^2 - 4) which is part of the table of standard integrals (not given anymore)?

$\noindent Probably. The answer to that is $\ln \left|x +\sqrt{x^{2} -4}\right|$, and can be shown with a trig. substitution. (Those ones they got rid of similar to the one you wrote can all be evaluated using trig. substitutions, so there's no reason students couldn't be expected to do them still.)$

pikachu975 · Feb 14, 2017

InteGrand said:
$\noindent Probably. The answer to that is $\ln \left|x +\sqrt{x^{2} -4}\right|$, and can be shown with a trig. substitution. (Those ones they got rid of similar to the one you wrote can all be evaluated using trig. substitutions, so there's no reason students couldn't be expected to do them still.)$

I got x = sect
dx = sect*tant*dt

=> integral sect*tant*dt / 2sqrt(sec^2 t - 1)
= 1/2 integral sect*dt

How would we integrate sect*dt?

InteGrand · Feb 14, 2017

pikachu975 said:
I got x = sect
dx = sect*tant*dt

=> integral sect*tant*dt / 2sqrt(sec^2 t - 1)
= 1/2 integral sect*dt

How would we integrate sect*dt?

$\noindent It's a well-known result that $\int \sec t \, \mathrm{d}t = \ln \left|\sec t + \tan t\right|$. One way to see this is to multiply top and bottom of the integrand by $\sec t + \tan t$ to make it $\int \frac{\sec^{2}t + \sec t \tan t}{\sec t + \tan t }\,\mathrm{d}t$ and observe that the numerator becomes the derivative of the denominator.$

pikachu975 · Feb 14, 2017

InteGrand said:
$\noindent It's a well-known result that $\int \sec t \, \mathrm{d}t = \ln \left|\sec t + \tan t\right|$. One way to see this is to multiply top and bottom of the integrand by $\sec t + \tan t$ to make it $\int \frac{\sec^{2}t + \sec t \tan t}{\sec t + \tan t }\,\mathrm{d}t$ and observe that the numerator becomes the derivative of the denominator.$

Is multiplying top and bottom by sect + tant something you have to deduce/remember from past papers or is it something I haven't learnt yet further in integration?

InteGrand · Feb 14, 2017

pikachu975 said:
Is multiplying top and bottom by sect + tant something you have to deduce/remember from past papers or is it something I haven't learnt yet further in integration?

It's basically an ad hoc method. You can also find the integral of the secant function more conventionally by using the "t-formulae", but you'd then need to use some trig. identities to convert the resulting answer into the answer I wrote in the earlier post.

Incidentally, finding the integral of the secant function was crucial for geography and map-making in the seventeenth century (for producing accurate Mercator projections). This can be read about here: http://www.math.uconn.edu/~kconrad/math1132s10/secantintegral.pdf . Here's the Wikipedia article on that integral: https://en.wikipedia.org/wiki/Integral_of_the_secant_function . (Finding that integral was one of the "outstanding open problems of the mid-seventeenth century".)

Integration Question (1 Viewer)

pikachu975

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