# Thread: Integration Question

1. ## Integration Question

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

2. ## Re: Integration Question

Originally Posted by pikachu975
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
$\noindent By \textbf{inspection}, the answer is -5\sqrt{4-x^{2}} + \sin^{-1}\left(\frac{x}{2}\right)+c.$

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

3. ## Re: Integration Question

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

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

4. ## Re: Integration Question

Originally Posted by pikachu975
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*}

5. ## Re: Integration Question

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

\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)?

6. ## Re: Integration Question

Originally Posted by pikachu975
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)?
$\noindent Probably. The answer to that is \ln \left|x +\sqrt{x^{2} -4}\right|, and can be shown with a trig. substitution. (Those ones they got rid of similar to the one you wrote can all be evaluated using trig. substitutions, so there's no reason students couldn't be expected to do them still.)$

7. ## Re: Integration Question

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

8. ## Re: Integration Question

Originally Posted by pikachu975
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?
$\noindent It's a well-known result that \int \sec t \, \mathrm{d}t = \ln \left|\sec t + \tan t\right|. One way to see this is to multiply top and bottom of the integrand by \sec t + \tan t to make it \int \frac{\sec^{2}t + \sec t \tan t}{\sec t + \tan t }\,\mathrm{d}t and observe that the numerator becomes the derivative of the denominator.$

9. ## Re: Integration Question

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

10. ## Re: Integration Question

Originally Posted by pikachu975
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/m...ntintegral.pdf . Here's the Wikipedia article on that integral: https://en.wikipedia.org/wiki/Integr...ecant_function . (Finding that integral was one of the "outstanding open problems of the mid-seventeenth century".)

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