Verified answer

1) ∫(t)*{sec²(8t)} dt [As derivative of t will become constant on differentiation, as well integral of sec² is known, let us take f(t), that is 1st function = t and g(t), that is 2nd function = sec²(8t).

==> f'(t) = 1; ∫g(t) = tan(8t)/8

2) Hence, ∫(t)*{sec²(8t)} dt = (t)*{tan(8t)}/8 - ∫(1)*{tan(8t)}/8 dt

= (t/8)*{tan(8t)} - (1/8)∫{tan(8t)}dt

= (t/8)*{tan(8t)} - (1/8)*[lnIsec(8t)I/8] + C

= (t/8)*{tan(8t)} - (1/64)*[lnIsec(8t)I] + C

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∫ t sec²(8t) dt = uv - ∫ v du

Let u = t

du = dt

Let dv = sec²(8t) dt

v = 1/8 tan(8t)

t(1/8 tan(8t)) - ∫ (1/8 tan(8t)) dt)

1/8 t tan(8t) - 1/8 ∫ tan(8t) dt + C

1/8 (t tan(8t) - ∫ tan(8t) dt) + C

1/8 (t tan(8t) + 1/8 ln|cos(8t)|) + C

1/64 (64t tan(8t) + ln|cos(8t)|) + C