Final answer:
To evaluate the integral, we can use the linearity property of integrals. We can integrate each term separately and then add the results together. The final result is t i + (1/3) tan³ t i + t i + (1/7) (t⁸ + t⁷) j + (1/9) t⁹ ln t k - (1/81) t⁹ k + C.
Step-by-step explanation:
To evaluate the integral ∫ (sec² t i + t(t² + 1)⁶ j + t⁸ ln t k) dt, we can use the linearity property of integrals. We can integrate each term separately and then add the results together.
For the first term, ∫ sec² t i dt, we can use the trigonometric identity sec² t = 1 + tan² t, which gives us ∫ (1 + tan² t) i dt. The integral of 1 is t, and the integral of tan² t is (1/3) tan³ t + t. So, the result for the first term is t i + (1/3) tan³ t i + t i.
Similarly, for the second term, we have ∫ t(t² + 1)⁶ j dt. We can distribute the t and apply the power rule of integration to get (1/7) (t⁸ + t⁷) j.
Finally, for the third term, ∫ t⁸ ln t k dt, we can use integration by parts. Let u = ln t and dv = t⁸ dt. Then, du = (1/t) dt and v = (1/9) t⁹. Applying the integration by parts formula, we get ∫ t⁸ ln t k dt = (1/9) t⁹ ln t - (1/9) ∫ t⁸ dt = (1/9) t⁹ ln t - (1/81) t⁹ k.
Adding all the terms together, the final result of the integral is t i + (1/3) tan³ t i + t i + (1/7) (t⁸ + t⁷) j + (1/9) t⁹ ln t k - (1/81) t⁹ k + C, where C is the constant of integration.