# Question Video: Finding the Integration of a Function Involving Trigonometric Functions Mathematics

Determine β« β8 sec 7π₯(β4 cosΒ² 7π₯ + 6 tan 7π₯) dπ₯.

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### 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₯.

Letβs 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 πΆ. And the integral of sec π₯ tan π₯ with respect to π₯ is equal to sec π₯ plus πΆ.

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 πΆ one minus 48 over seven sec π’ plus a constant of integration πΆ two. We can combine the two constants of integration into a single arbitrary constant πΆ. 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 πΆ.