What is the integral of tsec^2(8t) ?

The integral of tsec^2(8t) is 1/2 t^2sec^2(8t)-1/64 (32t^2tan^2(8t)-8ttan(8t)-ln|cos(8t)|+32t^2)+C

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integral of tsec^2(8t)

\int t sec^{2} (8 t) d t

\frac{1}{2} t^{2} sec^{2} (8 t) - \frac{1}{64} (32 t^{2} tan^{2} (8 t) - 8 t tan (8 t) - ln ∣ cos (8 t) ∣ + 32 t^{2}) + C

Solution steps

\int t sec^{2} (8 t) d t

Rewrite using trig identities

= \int t (1 + tan^{2} (8 t)) d t

Apply Integration By Parts

: \frac{1}{2} t^{2} sec^{2} (8 t) - \int 8 t^{2} sec^{2} (8 t) tan (8 t) d t

= \frac{1}{2} t^{2} sec^{2} (8 t) - \int 8 t^{2} sec^{2} (8 t) tan (8 t) d t

\int 8 t^{2} sec^{2} (8 t) tan (8 t) d t = \frac{1}{64} (32 t^{2} tan^{2} (8 t) - 8 t tan (8 t) - ln ∣ cos (8 t) ∣ + 32 t^{2})

= \frac{1}{2} t^{2} sec^{2} (8 t) - \frac{1}{64} (32 t^{2} tan^{2} (8 t) - 8 t tan (8 t) - ln ∣ cos (8 t) ∣ + 32 t^{2})

Add a constant to the solution

= \frac{1}{2} t^{2} sec^{2} (8 t) - \frac{1}{64} (32 t^{2} tan^{2} (8 t) - 8 t tan (8 t) - ln ∣ cos (8 t) ∣ + 32 t^{2}) + C

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What is the integral of tsec^2(8t) ?
The integral of tsec^2(8t) is 1/2 t^2sec^2(8t)-1/64 (32t^2tan^2(8t)-8ttan(8t)-ln|cos(8t)|+32t^2)+C