### Video Transcript

Determine the indefinite integral
of negative eight sec seven π₯ multiplied by negative four cos squared seven π₯ plus
six tan seven π₯ with respect to π₯.

Letβs begin by distributing the
parentheses in the integrand. The integrand becomes 32 sec seven
π₯ cos squared seven π₯ minus 48 sec seven π₯ tan seven π₯. Now since sec of seven π₯ is equal
to one over cos of seven π₯, we can replace the first term with 32 cos squared seven
π₯ over cos of seven π₯. We can cancel a factor of cos seven
π₯ from the numerator and denominator, so the first term simplifies to 32 cos seven
π₯. And now we need to determine the
indefinite integral of 32 cos seven π₯ minus 48 sec seven π₯ tan seven π₯ with
respect to π₯.

The first term in the integrand
involves a cosine function, and the second term involves a product of a secant and
tangent function each with the same argument of seven π₯. We will therefore need to recall
the following standard integrals. The integral of cos π₯ with respect
to π₯ is equal to sin π₯ plus πΆ. And the integral of sec π₯ tan π₯
with respect to π₯ is equal to sec π₯ plus πΆ.

However, before we can apply these
formulas to solve the given indefinite integral, we need to modify the argument of
the trigonometric functions. To do this, we can perform a
substitution. We let π’ equal seven π₯, which in
turn implies that dπ’ by dπ₯ is equal to seven. Now dπ’ by dπ₯ is not a fraction,
but we can treat it a little like one. So equivalently one-seventh dπ’ is
equal to dπ₯. Performing the substitution, we
obtain the indefinite integral of 32 cos π’ minus 48 sec π’ tan π’ one-seventh
dπ’.

We can simplify this by splitting
the integrand and taking each constant factor out the front to give 32 over seven
multiplied by the indefinite integral of cos π’ dπ’ minus 48 over seven multiplied
by the indefinite integral of sec π’ tan π’ dπ’. Applying the standard results, we
obtain 32 over seven sin π’ plus a constant of integration πΆ one minus 48 over
seven sec π’ plus a constant of integration πΆ two. We can combine the two constants of
integration into a single arbitrary constant πΆ. Finally, we need to reverse the
substitution by replacing π’ with seven π₯. And so we obtain our final answer,
which is 32 over seven sin seven π₯ minus 48 over seven sec seven π₯ plus πΆ.