Make a substitution to express the integrand as a rational function and then evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) 2 sec 2(t) tan2(t) + 16 tan(t) + 63 dt

integral of [2sec²(t) tan²(t) + 16tan(t) + 63] dt

= (⅔)tan³(t) - ln|cos(t)| + 63t + C

Step-by-step explanation:

We are required to evaluate the integral of

[2sec²(t) tan²(t) + 16tan(t) + 63] dt

after making a substitution to express the integral as a rational function.

2sec²(t) tan²(t) + 16tan(t) + 63

This expressions in this function can be integrated independently.

First, let us find

Integral of 2sec²(t) tan²(t) dt

Let u = tan(t)

du = sec²(t) dt

dt = du/sec²(t)

Integral of 2sec²(t) tan²(t) dt =

Integral of 2sec²(t). u² . du/sec²(t)

= integral of 2u²du

= (⅔)u³

Integral of 2sec²(t) tan²(t) dt = (⅔)tan³(t) + C1 (1)

Integral of 16tan(t)

= integral of sin(t)/cos(t) dt

= -ln|cos(t)| + C2 (2)

Integral of 63 dt

= 63t + C3(3)

Adding (1), (2) and (3), we have

integral of [2sec²(t) tan²(t) + 16tan(t) + 63] dt

= (⅔)tan³(t) - ln|cos(t)| + 63t + C

the answer

b. they are parallel and will never intersect.

bull

step-by-step explanation: