What is the integral of tsec^2(4t) ?

The integral of tsec^2(4t) is 1/2 t^2sec^2(4t)-1/16 (8t^2tan^2(4t)-4ttan(4t)-ln|cos(4t)|+8t^2)+C

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integral of tsec^2(4t)

\int t sec^{2} (4 t) d t

\frac{1}{2} t^{2} sec^{2} (4 t) - \frac{1}{16} (8 t^{2} tan^{2} (4 t) - 4 t tan (4 t) - ln ∣ cos (4 t) ∣ + 8 t^{2}) + C

Solution steps

\int t sec^{2} (4 t) d t

Rewrite using trig identities

= \int t (1 + tan^{2} (4 t)) d t

Apply Integration By Parts

: \frac{1}{2} t^{2} sec^{2} (4 t) - \int 4 t^{2} sec^{2} (4 t) tan (4 t) d t

= \frac{1}{2} t^{2} sec^{2} (4 t) - \int 4 t^{2} sec^{2} (4 t) tan (4 t) d t

\int 4 t^{2} sec^{2} (4 t) tan (4 t) d t = \frac{1}{16} (8 t^{2} tan^{2} (4 t) - 4 t tan (4 t) - ln ∣ cos (4 t) ∣ + 8 t^{2})

= \frac{1}{2} t^{2} sec^{2} (4 t) - \frac{1}{16} (8 t^{2} tan^{2} (4 t) - 4 t tan (4 t) - ln ∣ cos (4 t) ∣ + 8 t^{2})

Add a constant to the solution

= \frac{1}{2} t^{2} sec^{2} (4 t) - \frac{1}{16} (8 t^{2} tan^{2} (4 t) - 4 t tan (4 t) - ln ∣ cos (4 t) ∣ + 8 t^{2}) + C

y = x^{3} (2 x - 1)^{7}

\int x^{8} (x^{9} - 10)^{7} d x

\frac{\partial}{\partial y} (- 5 y z^{4})

\frac{d}{d x} (ln (\frac{4}{x ^{3}}))

\frac{\partial}{\partial x} (\frac{cos ( 2 y )}{e ^{2 x}})

What is the integral of tsec^2(4t) ?
The integral of tsec^2(4t) is 1/2 t^2sec^2(4t)-1/16 (8t^2tan^2(4t)-4ttan(4t)-ln|cos(4t)|+8t^2)+C