Video Transcript
Determine the indefinite integral
of two multiplied by csc of seven 𝑥 with respect to 𝑥.
Now, this is a really tricky
function to integrate. It requires us to really know our
derivatives of trigonometric functions. Now, since we’re trying to
determine the indefinite integral of two csc of seven 𝑥, let’s write down what the
derivative of csc of seven 𝑥 is. It is equal to negative seven csc
of seven 𝑥 multiplied by cot of seven 𝑥. Now what we’re aiming to do at this
stage is to try and get our integrand in the form of 𝑎 multiplied by 𝑓 prime of 𝑥
over 𝑓 of 𝑥. Since we know how to integrate this
and in order to try and get our integral in this form. We’re going to try and multiply the
integrand by a fraction consisting of a function over itself, which is of course
equal to one. The difficult part is trying to
find a function 𝑔 of 𝑥. So our integral will end up in the
form which we know how to integrate.
Let’s see what happens if we let 𝑔
of 𝑥 be equal to csc of seven 𝑥. We multiply our integrand by csc of
seven 𝑥 over csc of seven 𝑥. And we obtain the integral of two
csc squared of seven 𝑥 over csc of seven 𝑥 with respect to 𝑥. However, this is clearly not in the
form required. However, it does give us a hint
since we have a csc squared of seven 𝑥 in the numerator. And we in fact know another
trigonometric function which would differentiate to give us a multiple of csc
squared of seven 𝑥. And this is cot of seven 𝑥. The derivative of cot of seven 𝑥
is negative seven csc squared of seven 𝑥.
Now it’s at this stage where we
might start to be able to spot what our 𝑔 of 𝑥 is. When we multiply by 𝑔 of 𝑥 over
𝑔 of 𝑥 in our integrand, we are always gonna have that factor of csc of seven 𝑥,
which is originally in the integrand. If we factor out a factor of csc of
seven 𝑥 from our two differentials, then we’re left with a cot of seven 𝑥
multiplied by a constant and a csc of seven 𝑥 multiplied by the same constant. Now, this is very important since
the functions which we’re differentiating are csc of seven 𝑥 and cot of seven
𝑥.
And so maybe, for our next 𝑔 of
𝑥, we can try adding these two functions together. But first, let’s make a quick note
what the differential of the sum of these two functions is. Using the fact that the
differential of a sum of functions is equal to the sum of the differentials of the
functions. And by keeping the csc of seven 𝑥
term factored. We have that the differential of
csc of seven 𝑥 plus cot of seven 𝑥 is equal to csc of seven 𝑥 multiplied by
negative seven cot of seven 𝑥 minus seven csc of seven 𝑥.
So now we can try 𝑔 of 𝑥 being
equal to csc of seven 𝑥 plus cot of seven 𝑥. We have the integral of two csc of
seven 𝑥 multiplied by csc of seven 𝑥 plus cot of seven 𝑥 over csc of seven 𝑥
plus cot of seven 𝑥 with respect to 𝑥. And we can rewrite this with the
two csc of seven 𝑥 in the numerator. Now we can notice that our integral
is very nearly in the form which we require. If we let 𝑓 of 𝑥 be equal to 𝑔
of 𝑥, then we can see that the denominator of our integrand is 𝑔 of 𝑥. And if we compare the numerator
with the differential of 𝑔 of 𝑥, then we can see that it’s very, very similar. The only two slight differences is
that there’s this constant factor of two in the numerator of the integrand and a
factor of negative seven in the differential. And so we can come to the
conclusion that the numerator of our integrand is equal to two over negative seven
times by 𝑔 prime of 𝑥.
We can now more clearly see that
our integral is in fact of the form which we require. In our case, 𝑎 is equal to
negative two over seven. And 𝑓 of 𝑥 is equal to csc of
seven 𝑥 plus cot of seven 𝑥. And so we can apply the
formula. Here, we reach our solution, which
is that the indefinite integral of two csc of seven 𝑥 with respect to 𝑥 is equal
to negative two-sevenths times the natural logarithm of the absolute value of csc of
seven 𝑥 plus cot of seven 𝑥 plus 𝑐.
