Question

Evaluate the integral. (sec2(t) i + t(t2 + 1)3 j + t2 ln(t) k) dt

Answer 1

To evaluate the integral, we can use the linearity property of integration to evaluate each term separately. For sec^2(t), the integral is tan(t) + C. For t(t^2 + 1)^3, we can expand the expression and integrate each term. For t^2 ln(t), we can use integration by parts. Finally, we can add the results of each term to obtain the overall integral.

Step-by-step explanation:

To evaluate the integral

(sec2(t) i + t(t2 + 1)3 j + t2 ln(t) k) dt

We will use the linearity property of integration to evaluate each term separately.

The integral of sec2(t) dt is tan(t) + C, where C is the constant of integration.

The integral of t(t2 + 1)3 dt can be evaluated by expanding the expression using the binomial theorem and integrating each term.

The integral of t2 ln(t) dt can be solved using integration by parts.

Finally, we can add the results of each term to obtain the overall integral.

Answer 2

We have been given a vector valued function:

`sec^(2)(t)i+t(t^(2)+1)^(3)j+t^(2)ln(t)k`

In order to evaluate the integral of this vector valued function, we will integrate each component of the vector valued function.

`(\int sec^(2)(t)dt)i+(\int t(t^(2)+1)^(3)dt)j+(\int t^(2)ln(t)dt)k`

Upon integrating each of the components, we get:

`(\int sec^(2)(t)dt)i+(\int t(t^(2)+1)^(3)dt)j+(\int t^(2)ln(t)dt)k\\\\(tan(t)+c_(1))i+(((t^(2)+1)^(4))/(8)+c_(2))j+((t^3(3ln(t)-1))/(9)+c_(3))k`