What is the integral of sec^2(4t)(3-tan(4t))^3 ?

The integral of sec^2(4t)(3-tan(4t))^3 is 27/4 tan(4t)-27/8 tan^2(4t)+3/4 tan^3(4t)-1/16 tan^4(4t)+C

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integral of sec^2(4t)(3-tan(4t))^3

\int sec^{2} (4 t) (3 - tan (4 t))^{3} d t

\frac{27}{4} tan (4 t) - \frac{27}{8} tan^{2} (4 t) + \frac{3}{4} tan^{3} (4 t) - \frac{1}{16} tan^{4} (4 t) + C

Solution steps

\int sec^{2} (4 t) (3 - tan (4 t))^{3} d t

Expand

sec^{2} (4 t) (3 - tan (4 t))^{3} : 27 sec^{2} (4 t) - 27 sec^{2} (4 t) tan (4 t) + 9 sec^{2} (4 t) tan^{2} (4 t) - tan^{3} (4 t) sec^{2} (4 t)

= \int 27 sec^{2} (4 t) - 27 sec^{2} (4 t) tan (4 t) + 9 sec^{2} (4 t) tan^{2} (4 t) - tan^{3} (4 t) sec^{2} (4 t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 27 sec^{2} (4 t) d t - \int 27 sec^{2} (4 t) tan (4 t) d t + \int 9 sec^{2} (4 t) tan^{2} (4 t) d t - \int tan^{3} (4 t) sec^{2} (4 t) d t

\int 27 sec^{2} (4 t) d t = \frac{27}{4} tan (4 t)

\int 27 sec^{2} (4 t) tan (4 t) d t = \frac{27}{8} tan^{2} (4 t)

\int 9 sec^{2} (4 t) tan^{2} (4 t) d t = \frac{3}{4} tan^{3} (4 t)

\int tan^{3} (4 t) sec^{2} (4 t) d t = \frac{1}{16} tan^{4} (4 t)

= \frac{27}{4} tan (4 t) - \frac{27}{8} tan^{2} (4 t) + \frac{3}{4} tan^{3} (4 t) - \frac{1}{16} tan^{4} (4 t)

Add a constant to the solution

= \frac{27}{4} tan (4 t) - \frac{27}{8} tan^{2} (4 t) + \frac{3}{4} tan^{3} (4 t) - \frac{1}{16} tan^{4} (4 t) + C

What is the integral of sec^2(4t)(3-tan(4t))^3 ?
The integral of sec^2(4t)(3-tan(4t))^3 is 27/4 tan(4t)-27/8 tan^2(4t)+3/4 tan^3(4t)-1/16 tan^4(4t)+C