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