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25 April, 19:50

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

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  1. 25 April, 23:27
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    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
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