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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

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

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

Add a constant to the solution

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

\int_{- \infty}^{4} x e^{\frac{x}{2}} d x

y^{^{'}} = - y - 2 x

\int x (1 - 3 x^{2})^{4} d x

ln (x) \cdot x

\frac{\partial}{\partial u} (4 sin (u) cos (v))

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