Integration Substitution Techniques for quadratic expressions under square roots

[12Element]

Homework Statement

Integration by substitution technique for quadratic equation in square root.

Relevant Equations

intSqrt(q+px+X^2)dx

Hi,
With respect to the techniques mentioned in point 2 and 3:

Can someone explain or even better, post a link for an explanation or a videos showing the use of these two techniques.
Below excerpt shows problems 4 and 5 referenced in the above 2 points:

 

[PhDeezNutz]

Can you show problems 1-3?

As for an example of strategy 1

##\int \frac{1}{\sqrt[3]{4x+5}}\,dx##

Let ##z^3 = 4x+5## or ##z = \left( 4x+5 \right)^{\frac{1}{3}} ##

Take the derivative ##\frac{d}{dx}## of both sides (implicit differentiation)

##3z^2 \frac{dz}{dx} = 4##

##dx = \frac{3z^2}{4} dz##

So the integral now becomes

## \int \left(\frac{1}{\sqrt[3]{z^3}}\right) \cdot \left(\frac{3z^2}{4} \right) \,dz##

##= \int \left( \frac{3z^2}{4z} \right) \,dz##

##=\frac{3}{4}\int z \,dz = \frac{3z^2}{8} = \frac{3}{8} \left( \left( 4x+5 \right)^{\frac{1}{3}}\right)^2##

## = \frac{3}{8} \left(4x + 5 \right)^{\frac{2}{3}} + C ##

 

[12Element]

Thanks, but my question is with regard to point 2 and 3. I am well aware about how to use the technique of point 1. Also, problem 1-3 are related to point 1.

 

[PhDeezNutz]

Of course a constant of integration, and I edited my post to include equals signs.

 

[Mark44]

Thanks, but my question is with regard to point 2 and 3. I am well aware about how to use the technique of point 1. Also, problem 1-3 are related to point 1.

There is an apparent typo in the text you're quoting from.
Right after problem 4, it says:

Let ##x^2 = x + 2 = (z - x)^2## Then

That first '=' should be '+'.

There's also a typo in the next line, in the equation for dx.

$$dx = \frac{2(z^2 + z + z)}{(1 + 2z)^2}$$

That 2nd z term in the numerator should be '2', not 'z'.
With that many typos, the work becomes somewhat suspect.

If you're having difficulties with a specific problem, please post that problem and we can help you with it.

 

[12Element]

Thanks for taking the time to answer my queries, I really appreciate it.

You are right, there are quite a few typos in the text. I guess my questions would be if:

$$ x^2 + x + 2 = (z-x)^2 $$ then: $$ \sqrt{x^2 + x + 2} = (z-x) $$ how did the author get: $$ \sqrt{x^2 + x + 2} = \frac {(z^2 + z + 2)} {1 + 2z} $$ also, how did he obtain x: $$ x = \frac {z^2 - 2} {1+2z} $$

The way I would attempt to solve it, is very tedious. Whilst the technique/substitution used by the author above seems to be very succinct and ingenious (if it works though).

My way would be as per the following (I probably made quite few errors along the way however, please let me know if the methodology used is wrong in principle, or if a better substitution can be used):

First complete the square of the quadratic in the square root:

$$ \int \frac {dx}{x + \sqrt{\left(x+\frac{1}{2}\right)^2 + \frac{7}{4}}} \,dx $$ then, set u to be: $$ u=x+\frac{1}{2} $$ hence, x would be: $$ x=u-\frac{1}{2} $$ therefore: $$ \int \frac {du}{u-\frac{1}{2} \sqrt{u^2 + \frac{7}{4}}} \,du $$ now replace u and du with: $$ \frac{\sqrt{7}}{2}\tan\theta = u $$ and $$ du =\frac{\sqrt{7}}{2} \sec^{2}\theta \, d\theta $$ Therefore, $$ \int \frac {\frac{\sqrt{7}}{2} \sec^{2}\theta}{\left(\frac{\sqrt{7}}{2}\tan\theta-\frac{1}{2} \right) \frac{\sqrt{7}}{2} \sec\theta} \,d\theta $$ simplifying: $$ =\int \frac {\sec\theta}{\frac{\sqrt{7}}{2}\tan\theta-\frac{1}{2}} \,d\theta $$ further simplifying: $$ =\int \frac {d\theta}{\frac{\sqrt{7}}{2}\sin\theta-\frac{1}{2}\cos\theta} $$ from here the only substitution I can think of is to set: $$ \tan\frac{\theta}{2}=t \ $$ then, using the double angle trig identity we have: $$ \cos\theta= \frac{1-t^2}{1+t^2} \text{, } \sin\theta= \frac{2t}{1+t^2} $$ differentiating the inverse tan, we get: $$ d\theta= \frac {2}{1+t^2} dt $$ substituting in our integral, we have: $$ =\int \frac {\frac {2}{1+t^2} dt}{\frac{\sqrt{7}}{2}\frac{2t}{1+t^2}-\frac{1}{2}\frac{1-t^2}{1+t^2}} $$ Simplifying: $$ =\int \frac {4dt}{2\sqrt{7}t-1+t^2} $$ Solving the quadratic, we have: $$ =\int \frac {4dt}{\left(t+\sqrt{7}+2\sqrt{2}\right)\left(t+\sqrt{7}+2\sqrt{2}\right)} $$ and then using partial fraction decomposition, we have: $$ =\frac{1}{\sqrt{2}}\int \frac {dt}{t+\sqrt{7}-2\sqrt{2}} - \frac{1}{\sqrt{2}}\int \frac {dt}{t+\sqrt{7}+2\sqrt{2}} $$ finally, integrating: $$ =\frac{1}{\sqrt{2}}\ln\left|t+\sqrt{7}-2\sqrt{2} \right| - \frac{1}{\sqrt{2}}\ln\left|t+\sqrt{7}+2\sqrt{2} \right| $$ simplifying: $$ =\frac{1}{\sqrt{2}}\ln\left|\frac {{t+\sqrt{7}-2\sqrt{2}}}{t+\sqrt{7}+2\sqrt{2}} \right| $$ remembering that: $$ \theta = \tan^{-1}\left( \frac{2u}{\sqrt{7}}\right) $$ substituting: $$ =\frac{1}{\sqrt{2}}\ln\left|\frac {{\tan\frac{\theta}{2}+\sqrt{7}-2\sqrt{2}}}{\tan\frac{\theta}{2}+\sqrt{7}+2\sqrt{2}} \right| $$ remembering that: $$ \theta = \tan^{-1}\left( \frac{2u}{\sqrt{7}}\right) $$ substituting: $$ =\frac{1}{\sqrt{2}}\ln\left|\frac {{\tan\left(\frac{\tan^{-1}\left( \frac{2u}{\sqrt{7}}\right)}{2}\right)+\sqrt{7}-2\sqrt{2}}}{\tan\left(\frac{\tan^{-1}\left( \frac{2u}{\sqrt{7}}\right)}{2}\right)+\sqrt{7}+2\sqrt{2}} \right| $$ finally, remembering that: $$ x=u-\frac{1}{2} $$ substituting: $$ =\frac{1}{\sqrt{2}}\ln\left|\frac {{\tan\left(\frac{\tan^{-1}\left( \frac{2x+1}{\sqrt{7}}\right)}{2}\right)+\sqrt{7}-2\sqrt{2}}}{\tan\left(\frac{\tan^{-1}\left( \frac{2x+1}{\sqrt{7}}\right)}{2}\right)+\sqrt{7}+2\sqrt{2}} \right| $$

Again, I am sure I made mistakes along the way, however is my method correct in principle. Also, do you have any other suggestion or can figure out what the text author is doing?

 

[DrClaude]

Thanks for taking the time to answer my queries, I really appreciate it.

You are right, there are quite a few typos in the text. I guess my questions would be if:

$$ x^2 + x + 2 = (z-x)^2 $$ then: $$ \sqrt{x^2 + x + 2} = (z-x) $$

##x## appears on both sides of the equality, so this is incomplete.

how did the author get: $$ \sqrt{x^2 + x + 2} = \frac {(z^2 + z + 2)} {1 + 2z} $$ also, how did he obtain x: $$ x = \frac {z^2 - 2} {1+2z} $$

Consider first the latter: ##x## is obtained by solving
$$
x^2 + x + 2 = (z-x)^2
$$
then ##\sqrt{x^2 + x + 2}## is obtained by substituting the result for ##x## in ##z-x##.

 

[12Element]

Thanks a lot, all is clear now.

 

[Mark44]

I guess my questions would be if:
## x^2 + x + 2 = (z-x)^2##
then: ## \sqrt{x^2 + x + 2} = (z-x) ##

That would be 1) unhelpful, because you still have x on both sides (as already pointed out), and 2) incorrect, because ##\sqrt{(z-x)^2} \ne z - x##. The correct expression on the right side would be |z - x|. For example, ##\sqrt{ (-4)^2} \ne -4##.

 

[PhDeezNutz]

Recently @fresh_42 wrote an insight article "The Art of Integration". I think Euler's second substitution applies to this problem.

https://www.physicsforums.com/insights/the-art-of-integration/#Euler-Substitutions