MHB Exam Prep Part 1: Distinguishing Rules & Techniques

[bmanmcfly]

So, I'm gearing up for a final exam her, and while I handled learning the integration of trig,log and exponential functions...
Nowthat I've got a good grasp of all these rules I'm getting lost in the sheer vastness of what I need to be aware of...

For example, how do you, in general distinguish between the different ways to swap x for du and u? I mean, sometimes you will see sweeping out the x, sometimes the ax, sometimes the x2^2-a, or other times you will swap the whole trig function, other times it takes just the expression within (the angle).

So, how does one distinguish?

On the same note, how do we decide the difference between inverse trig, general power and log rules?

As far as this all goes, I understand the rules, but in the vastness of possibilities I find my use o these things approaching chance.

 

[alyafey22]

Sorry , I cannot understand what you are asking ? , can you give an example :) ?

 

[bmanmcfly]

Sorry , I cannot understand what you are asking ? , can you give an example :) ?

This was more general question... Like, for for example, you might see [math]\int e^xdx[/math]
Where you will substitute [math]u=e^x anddu=e^xdx[/math], where in another similar example you will substitute just the x portion.

Another example that had me good was [math]\int \frac{sin^2(x)dx}{1-\cos(x)}[/math], where it turned out took some wild substitutions).

On the same page is when you have [math]\frac{dx}{x^2-1}[/math], [math]\frac{xdx}{x^2-1}[/math], etc... How to recognize when convert to sin, a log function or Or to use the general power rule?

Its just, in my practis, when given the choice I make the wrong choice more often then not.Same deal with integrating by parts.

 

[MarkFL]

To address your examples:

1.) Since $$\frac{d}{dx}\left(e^x \right)=e^x$$ then we must have:

$$\int e^x\,dx=e^x+C$$

2.) $$\int\frac{\sin^2(x)}{1-\cos(x)}\,dx$$

I would rewrite the integrand using a Pythagorean identity on the numerator, then factor as cancel as follows:

$$\frac{\sin^2(x)}{1-\cos(x)}=\frac{1-\cos^2(x)}{1-\cos(x)}=\frac{(1+\cos(x))(1-\cos(x))}{1-\cos(x)}=1+\cos(x)$$

Now it is easily integrable.

3.) $$\int\frac{dx}{x^2-1}$$

Here we should observe that partial fraction decomposition will work. After applying this (I used the Heaviside cover-up method) we get:

$$\frac{1}{2}\int\frac{1}{x-1}-\frac{1}{x+1}\,dx$$

Now it is easily integrable.

4.) $$\int\frac{x}{x^2-1}\,dx$$

Here we should spot that the numerator is a multiple of the derivative of the denominator, and so a $u$-substitution will work:

$$u=x^2-1\,\therefore\.du=2x\,dx$$ and we have:

$$\frac{1}{2}\int\frac{du}{u}$$

Now it is easily integrable.

As far as integration by parts, I use the LIATE mnemonic to decide which is to be $u$ and $dv$:

Integration by parts - Wikipedia, the free encyclopedia

 

[alyafey22]

This was more general question... Like, for for example, you might see [math]\int e^xdx[/math]
Where you will substitute [math]u=e^x anddu=e^xdx[/math], where in another similar example you will substitute just the x portion.

Another example that had me good was [math]\int \frac{sin^2(x)dx}{1-\cos(x)}[/math], where it turned out took some wild substitutions).

On the same page is when you have [math]\frac{dx}{x^2-1}[/math], [math]\frac{xdx}{x^2-1}[/math], etc... How to recognize when convert to sin, a log function or Or to use the general power rule?

Its just, in my practis, when given the choice I make the wrong choice more often then not.Same deal with integrating by parts.

In solving integrals , we might look at certain techniques to solve certain integrals but that doesn't mean that it is the only way to solve it . First , you have to know that substitution is only to simplify the integral , it still can be solved without substitution but we do it because we want to get a simplified expression that looks "easier" . Sometimes we do little simplifications before we proceed .
Here are some tips that you can follow :

-Try to simplify the expression . You might be able to factorize and cancel some terms.
-Think about a substitution that will make the problem easier to solve , in general you can obtain this by practice .
-Think about elementary integration techniques you studied in calculus [integration by parts , integration by partial fractions , integration by trig-substitution ...]

Before you start a substitution look for the term you want to substitute and try to see whether the derivative is in the expression or not ?

For example see the integration $$\int \frac{\ln (x)}{x} \, dx $$ here we know that the derivative of $$\ln(x) $$ is $$\frac{1}{x}$$ since there exists no other terms , you can do the following substitution $$\ln(x)= u$$

If ,on the other hand, you see a square root with some terms inside we usually think about trig-substitution. For example , look at $$\int \frac{dx}{ \sqrt{1-x^2}}$$ make the sub $$x=\sin(u)$$ . You must see your calculus book for other similar patterns.

Most of the time ,we use integration by parts when there are a polynomial multiplied by an exponential or trigonometric function .For example see the integration $$\int x \, e^x \, dx$$ , and we want to reduce this to something we know how to integrate so we differentiate $x$ and integrate $e^x$ .

We use partial fractions to discuss integration of fractions of polynomials , look at the example Mark provided .There are several techniques regarding this , you have to look at every single example and practice more .

 

[Petrus]

To address your examples:

1.) Since $$\frac{d}{dx}\left(e^x \right)=e^x$$ then we must have:

$$\int e^x\,dx=e^x+C$$

2.) $$\int\frac{\sin^2(x)}{1-\cos(x)}\,dx$$

I would rewrite the integrand using a Pythagorean identity on the numerator, then factor as cancel as follows:

$$\frac{\sin^2(x)}{1-\cos(x)}=\frac{1-\cos^2(x)}{1-\cos(x)}=\frac{(1+\cos(x))(1-\cos(x))}{1-\cos(x)}=1+\cos(x)$$

Now it is easily integrable.

3.) $$\int\frac{dx}{x^2-1}$$

Here we should observe that partial fraction decomposition will work. After applying this (I used the Heaviside cover-up method) we get:

$$\frac{1}{2}\int\frac{1}{x-1}-\frac{1}{x+1}\,dx$$

Now it is easily integrable.

4.) $$\int\frac{x}{x^2-1}\,dx$$

Here we should spot that the numerator is a multiple of the derivative of the denominator, and so a $u$-substitution will work:

$$u=x^2-1\,\therefore\.du=2x\,dx$$ and we have:

$$\frac{1}{2}\int\frac{du}{u}$$

Now it is easily integrable.

As far as integration by parts, I use the LIATE mnemonic to decide which is to be $u$ and $dv$:

Integration by parts - Wikipedia, the free encyclopedia

I want to add some part for the partial fraction, you probably know this but I would like to add that always remember to look if the degree at bottom is lower then the degree at top and then you can do long polynom division, An example of this is
$$\int \frac{x^4-2x^2+4x+1}{x^3-x^2-x+1}dx$$ first use long polynom division then do partial fraction, if you get stuck on this one we will help you!

Regards,
$$|\pi\rangle$$

 

[bmanmcfly]

To address your examples:

1.) Since $$\frac{d}{dx}\left(e^x \right)=e^x$$ then we must have:

$$\int e^x\,dx=e^x+C$$

This example was too simple... The question would be appropriate as when, or what would you look for to substitute the ^ax or substituting the whole e^ax?

2.) $$\int\frac{\sin^2(x)}{1-\cos(x)}\,dx$$

I would rewrite the integrand using a Pythagorean identity on the numerator, then factor as cancel as follows:

$$\frac{\sin^2(x)}{1-\cos(x)}=\frac{1-\cos^2(x)}{1-\cos(x)}=\frac{(1+\cos(x))(1-\cos(x))}{1-\cos(x)}=1+\cos(x)$$

Now it is easily integrable.

Ok, I thought that would be a valid substitution, and much simpler than the alternative...

3.) $$\int\frac{dx}{x^2-1}$$

Here we should observe that partial fraction decomposition will work. After applying this (I used the Heaviside cover-up method) we get:

$$\frac{1}{2}\int\frac{1}{x-1}-\frac{1}{x+1}\,dx$$

Now it is easily integrable.

4.) $$\int\frac{x}{x^2-1}\,dx$$

Here we should spot that the numerator is a multiple of the derivative of the denominator, and so a $u$-substitution will work:

$$u=x^2-1\,\therefore\.du=2x\,dx$$ and we have:

$$\frac{1}{2}\int\frac{du}{u}$$

Now it is easily integrable.

As far as integration by parts, I use the LIATE mnemonic to decide which is to be $u$ and $dv$:

Integration by parts - Wikipedia, the free encyclopedia

Thanks, this was all helpful...