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

Determine ∫ βˆ’8 sec 7π‘₯(βˆ’4 cosΒ² 7π‘₯ + 6 tan 7π‘₯) dπ‘₯.

02:44

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‘₯.

Let’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 𝐢. And the integral of sec π‘₯ tan π‘₯ with respect to π‘₯ is equal to sec π‘₯ plus 𝐢.

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 𝑒 equal seven π‘₯, which in turn implies that d𝑒 by dπ‘₯ is equal to seven. Now d𝑒 by dπ‘₯ is not a fraction, but we can treat it a little like one. So equivalently one-seventh d𝑒 is equal to dπ‘₯. Performing the substitution, we obtain the indefinite integral of 32 cos 𝑒 minus 48 sec 𝑒 tan 𝑒 one-seventh d𝑒.

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 𝑒 d𝑒 minus 48 over seven multiplied by the indefinite integral of sec 𝑒 tan 𝑒 d𝑒. Applying the standard results, we obtain 32 over seven sin 𝑒 plus a constant of integration 𝐢 one minus 48 over seven sec 𝑒 plus a constant of integration 𝐢 two. We can combine the two constants of integration into a single arbitrary constant 𝐢. Finally, we need to reverse the substitution by replacing 𝑒 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 𝐢.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.