What is the integral of 8tan^2(x)+tan^4(x) ?

The integral of 8tan^2(x)+tan^4(x) is -7x+1/3 tan^3(x)+7tan(x)+C

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integral of 8tan^2(x)+tan^4(x)

\int 8 tan^{2} (x) + tan^{4} (x) d x

- 7 x + \frac{1}{3} tan^{3} (x) + 7 tan (x) + C

Solution steps

\int 8 tan^{2} (x) + tan^{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 8 tan^{2} (x) d x + \int tan^{4} (x) d x

\int 8 tan^{2} (x) d x = 8 (- x + tan (x))

\int tan^{4} (x) d x = \frac{1}{3} tan^{3} (x) + x - tan (x)

= 8 (- x + tan (x)) + \frac{1}{3} tan^{3} (x) + x - tan (x)

Simplify

8 (- x + tan (x)) + \frac{1}{3} tan^{3} (x) + x - tan (x) : - 7 x + \frac{1}{3} tan^{3} (x) + 7 tan (x)

= - 7 x + \frac{1}{3} tan^{3} (x) + 7 tan (x)

Add a constant to the solution

= - 7 x + \frac{1}{3} tan^{3} (x) + 7 tan (x) + C

y^{2} = 4 x + 4, y = 4 x + 16

\frac{d}{d x} (\frac{x - 3}{3 + sin ( x )})

y = \frac{3}{x}, at P (0.5, 6)

\frac{d y}{d x} = - 3 x y^{3}

x \to 1 lim (x^{2} - 3)

What is the integral of 8tan^2(x)+tan^4(x) ?
The integral of 8tan^2(x)+tan^4(x) is -7x+1/3 tan^3(x)+7tan(x)+C