Welcome to the first of several videos on integration involving inverse trigonometric functions
Let's go ahead and take a look at our integration rules
in order to apply these properly I think pattern recognition is probably the most important skill to have
notice its only one that does not have a square root in the denominator
so if we don't have a square root we're going to be dealing with arc tangent
and then if it does have a square root notice that one of them has noticed the formula for arc sine
is a squared minus u squared
and the formula for arc secant is u squared minus a squared
and this formula also has u outside the square root
now you might be asking where are the formulas involving arc cosine, arc cotangent, and arc cosecant
remember when were doing derivatives
this first column and the second column were just opposite of one another
so it's pretty standard for textbooks when integrating we use
these three inverse trig functions rather than these three
because if we did integrate this with respect to u
instead of saying arc cosine u we could say negative arc sine u
Let's go and take a look at our examples
again the first thing we want to do is to be able to recognize which formula we are going to apply
and the key is a denominator we have the square root of 16 minus x squared
if we look at these three formulas
that pattern fits the first one
Let's go ahead and rewrite this and see if we can make it fit this pattern
Let's first factor the two out
so we have dx in the numerator
and for our denominator
we want the squared root of a squared minus u squared
so we can rewrite 16 as four squared
and u squared would be x squared
so what this tells us is that u is equal to x and a is equal to four
now the next thing we should recognize is that all of these formulas are in terms of u
so it's important that we find differential u to make sure we do not have to perform a substitution
in this case du would be equal to one dx or just dx
so no substitution is required
so now we can apply the formula
so the anti derivative would be two times
arc sine u over a
well u is x
and a is four
plus C
So this one was pretty straight forward let's go and take a look at another one
again step one is to identify the correct formula
notice our denominator does not have a square root so were looking at the second formula now
so our anti derivative should involve arc tangent
Let's go ahead and see if we can identify the values of a and u
by making this fit the pattern so our denominator needs to be a squared plus u squared
well if we let a squared equals five
that means a would be the square root of five
so we can rewrite this as a square root of five squared
plus
for u squared
if u squared is equal to 16x squared u would be 4x
so if we square 4x we will get 16x squared
so this tells us that a
is equal to the square root of five
and u is equal to 4x
again this formula is in terms of u so it's important that we determine du
well du
would equal four dx
since our current integrand has just dx let's go ahead and divide both sides by four
so we have one fourth du
is equal to
dx
Let's go ahead and rewrite this in terms of u using a, u
and one fourth u equal dx
kay so dx is equal to one fourth du
the square root of five
is equal to a so we would have a squared
plus
and u is equal to four x so we do have u squared
so this fits the pattern that we need to apply the formula
so we're gonna have one fourth
times
this anti derivative which would be one over a so one over the square root of five
times arc tangent
of u over a plus C
again u is four x
and a square root of five
and of course we can rewrite this first fraction as one over four square root of five
Let's go ahead and do that
Let's go and take a look at one more
again once you stand out for this one is it has a square root
and we also have x outside the square root so for this integral we'll apply the third formula
now that we've identified our formula let's go ahead and see if we can make it fit the pattern
to identify a and u
so notice that u squared must equal nine x squared
so that means u would be three x
and then a squared must equal four so this would be minus
two squared
and we do have a little bit of a problem here
we can see that a is going to equal two
and u has to be equal to three x
but notice that we have to have a u outside the square root as well and we don't we only have x
since we do have to have a three here
we can multiply the top and bottom by three so this would be three x
as long we have three dx in the numerator
now notice that it does fit the pattern there's u there's u squared
and there's a squared
next we need to identify du
well du
would equal three
dx
if we look at our integrand we do have a three dx so all of this will be equal to du
Let's go ahead and take this and rewrite it in terms of u and a to make sure it fit the pattern
so three dx is equal to du
three x is equal to u
so this would be u squared
and a is equal to two so this would be minus a squared
again it's a perfect match
so now we can find the anti derivative
it would be equal to one over a or one half
times arc secant
of the absolute value of u over a
absolute value of three x
over two and then plus C
okay this is the first of several videos on
integrals involving inverse trigonometric functions
so I do recommend you watch part two