Contents of this Calculus 1 episode:

Integration by substitution, useful substitutions, dx/du, integration by parts, Linear expression under the radical, Non-linear expression under the radical.

INTEGRATION BY SUBSTITUTION

The main idea of integration by substitution is that we replace an expression by u, hoping that this way we may be able to solve the problem.

Let's see an example for this!

In such cases it is best to assign u to be the entire square root expression.

USEFUL SUBSTITUTIONS

So far so good, now we solve for x.

And presumably, we would later substitute that here.

But sadly, there is one more thing here.

Well, dx should be replaced, too, the following way:

We differentiate the expressed x with respect to u.

If what you see here is surprising to you, well, there is an explanation for everything.

Long-long time ago, people did not use the well-known f’ notation for differentiation. Instead, they used

Some people still use this notation.

Later the notations were simplified, but for some mysterious reason, here, at

integration by substitution, this ancient method was kept.

So, let's resign ourselves to this, the derivative of the expressed x is not the usual

but

Let's hope that nobody would lose sleep over this.

And now comes the substitution.

Well, this seems to be done.

Let's see another one like this.

Integration by substitution is also useful for dealing with ugly composite functions.

In such cases we usually assign u to the internal function.

Now, we integrate by parts.

Let's see another one like this.

We have to express x from this

using a little trick.

There is such a thing as:

so this one is done.

We only have to differentiate with respect to u.

And here comes the substitution.

Now, we integrate. The derivative of the denominator

happens to be the numerator.

There are more exciting substitutions, too.

They will appear in the next slideshow. But only for those who like such thrills.

Integration by substitution works so that we replace an expression by u, hoping that this way we may be able to solve the problem.

So far, we worked on easier substitutions, now we are ready for more difficult ones. Let's start by looking at some square root cases.

When the expression under the square root sign is linear, that is the more pleasant case, and it is best to substitute u for the entire square root expression.

If, however, the expression under the square root sign is non-linear, then we are dealing with an entirely different animal.

In such cases it doesn't always worth to try substitution.

It is quite suspicious that the derivative of the1 – x4 under the square root sign is almost the same as the numerator.

So, we do not substitute, but instead

use a method called S2.

But if, annoyingly instead of x3, there is x2 in the numerator,

then the previous method doesn't work.

So, in our despair, we opt for integration by substitution.

Here is the menu.

This is exactly what we need now.

Let's see where this takes us.

Now come where we differentiate x.

First, we differentiate the external function,

and then the internal.

So far, this does not look very encouraging. But let's see the substitution.

And now comes the main point.

We do this whole substitution business, because

and this way we can get rid of the square root sign.

There is also the added benefit of being able to simplify.

At the end we realize that this is the easiest integral of our lives.

All we have left to do is replacing u with whatever we called u at the beginning.

Wait, what did we call u?

Now we take the arcsinus of both sides.

This is the inverse of sinus, thus

Let's see another exciting case.

Here is where we get rid of the square root sign.

And now it would be good to integrate.

We need a bit of a trick for that.

This one again…

Then we do some cutting (T1)

Finally, we replace whatever we called u.

The next story is about substituting trigonometric functions.

Let's see a square root case in this j(t) internal function version.

In such cases it is best to assign t to the entire square root expression in the denominator.

So, , thus and .

This latter one will be the j(t) function,

in other words, and .

After completing the substitution, the problem looks like this:

Here, we have to substitute for t, and then it is done.

Calculus 1 episode