### Video Transcript

Determine the indefinite integral
of 27 sin π₯ plus 21 cos π₯ all over seven sin π₯ minus nine cos π₯ with respect to
π₯.

Here, we notice that the numerator
of the integrand looks as though it could be the differential of the
denominator. Since the differential of sin π₯
with respect to π₯ is equal to cos π₯ and the differential of negative cos π₯ with
respect to π₯ is equal to sin π₯. Now it looks as though the
numerator may differ from the differential of the denominator by a constant
factor. However, we do not know what this
factor is. We can try and find it by using a
substitution. Letβs let π’ be equal to the
denominator of the integrand, so that seven sin π₯ minus nine cos π₯. Now we can differentiate π’ with
respect to π₯. Using the fact that sine
differentiates to cos π₯ and negative cos π₯ differentiates to sin π₯. We obtain that dπ’ by dπ₯ is equal
to seven cos π₯ plus nine sin π₯. This gives us that dπ’ is equal to
nine sin π₯ plus seven cos π₯ dπ₯.

Now letβs rearrange our integral so
we can apply this substitution. We notice that we can factor out a
factor of three from our numerator. And this enables us to write our
integral as the integral of three over seven sin π₯ minus nine cos π₯ multiplied by
nine sin π₯ plus seven cos π₯ dπ₯. And so we can substitute π’ into
the denominator of our fraction. And we can substitute in dπ’ for
nine sin π₯ plus seven cos π₯ dπ₯. Giving us that it is equal to the
integral of three over π’ dπ’, which can be integrated to three multiplied by the
natural logarithm of the absolute value of π’ plus π. For our final step, we simply
substitute back in seven sin π₯ minus nine cos π₯ for π’. This gives us our solution, which
is three multiplied by the natural logarithm of the absolute value of seven sin π₯
minus nine cos π₯ plus π.