Video Transcript
Evaluate the integral of one
divided by the square root of five plus four 𝑥 minus 𝑥 squared with respect to
𝑥.
In this question, we’re asked to
evaluate a definite integral. And we can see this is in a form
which we don’t know how to integrate. This means we’re going to need to
manipulate our integral into a form which we can integrate. And there’s a few different things
we might want to try. For example, we could try factoring
our quadratic. And if we were to do this, we would
see it factors fully. It’s equal to negative one
multiplied by 𝑥 plus one times 𝑥 minus five. However, writing this quadratic in
this way doesn’t help us. For example, we can’t simplify the
square root, and we still don’t know how to evaluate this integral.
But this is not the only thing we
could’ve done to manipulate our quadratic. We also could’ve tried completing
the square, so we’ll try completing the square. First, we see that our coefficient
of 𝑥 squared is negative one. So remember this means we need to
multiply by negative one. The next step is usually to divide
our coefficient of 𝑥 by two. But remember, we’re multiplying
through by negative one. So in fact, we need to divide by
negative two. This gives us negative one
multiplied by 𝑥 minus two all squared. Now we can ask the question, what
happens if we were to distribute the exponent over our parentheses?
By either using the FOIL method or
binomial expansion, we get negative one multiplied by 𝑥 squared minus four 𝑥 plus
four, which if we distribute negative one over our parentheses, we get negative 𝑥
squared plus four 𝑥 minus four. And this is almost exactly what we
have in our denominator. However, we can see the constant in
the quadratic in our denominator is positive five. And to make our constant five, we
need to add nine to negative four. Therefore, if we add nine to
negative one times 𝑥 minus two all squared plus nine, we get the quadratic in our
denominator, five plus four 𝑥 minus 𝑥 squared. And we can then use this to rewrite
our integral.
Doing this, we get the integral
given to us in the question is equal to the integral of one divided by the square
root of negative one times 𝑥 minus two all squared plus nine with respect to
𝑥. And at first, it might be hard to
see how this helps us evaluate this integral. However, this is actually very
similar to an integral we know how to evaluate. We need to recall the following
integral result. For any real constant 𝑎, the
integral of one over the square root of 𝑎 squared minus 𝑥 squared with respect to
𝑥 is equal to the inverse sin of 𝑥 over 𝑎 plus the constant of integration
𝐶.
And we can see that our integrand
is almost in this form. First, we need to rearrange the two
terms inside of the square root in our denominator. Doing this, we get the following
expression. Next, we can see instead of
subtracting 𝑥 squared in our denominator, we’re instead subtracting 𝑥 minus two
all squared. So to use this result, we instead
need our variable squared in this position. We can do this by using a
substitution. We want to use the substitution 𝑢
is equal to 𝑥 minus two. If 𝑢 is equal to 𝑥 minus two,
we’ll need to differentiate both sides of this expression with respect to 𝑥. We get d𝑢 by d𝑥 is equal to
one.
And of course we know d𝑢 by d𝑥 is
not a fraction. However, when we’re using
integration by substitution, it can help to treat it a little bit like a
fraction. This gives us the equivalent
statement in terms of differentials, d𝑢 is equal to d𝑥. And we can now evaluate this
integral by using our substitution. We replace 𝑥 minus two in our
denominator with 𝑢, and because d𝑢 is equal to d𝑥, we’re now integrating with
respect 𝑢. This gives us the integral of one
over the square root of nine minus 𝑢 squared with respect to 𝑢. And this is now almost exactly in
the form we need for our integral rule.
All we need to do now is find the
value of the constant 𝑎. And to find this, we just rewrite
nine as three squared. Now we can see the value of our
constant 𝑎 is equal to three. This means we can apply our
integral rule with the variable 𝑥 equal to 𝑢. So by setting our value of 𝑎 equal
to three and applying our integral rule, we get the inverse sin of 𝑢 over three
plus our constant of integration 𝐶. However, remember, our original
integral is in terms of 𝑥, so we should give our answer in terms of 𝑥. We can do this by using our
substitution 𝑢 is equal to 𝑥 minus two. And by doing this, we get the
inverse sin of 𝑥 minus two all over three plus the constant of integration 𝐶,
which is our final answer.
Therefore, by completing the square
in our denominator, we were able to show the integral of one over the square root of
five plus four 𝑥 minus 𝑥 squared with respect to 𝑥 is equal to the inverse sin of
𝑥 minus two all over three plus the constant of integration 𝐶.
