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

The integral of 4(tan^2(x)+tan^4(x)) is 4/3 tan^3(x)+C

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

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

\frac{4}{3} tan^{3} (x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

= 4 \cdot \int - 1 + sec^{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

= 4 (- \int 1 d x + \int sec^{2} (x) d x + \int tan^{4} (x) d x)

\int 1 d x = x

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

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

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

Simplify

4 (- x + tan (x) + \frac{1}{3} tan^{3} (x) + x - tan (x)) : \frac{4}{3} tan^{3} (x)

= \frac{4}{3} tan^{3} (x)

Add a constant to the solution

= \frac{4}{3} tan^{3} (x) + C

y^{^{''}} - 6 y^{^{'}} + 5 y = 0

\frac{\partial}{\partial x} (5 - x^{2} + x y - 3 y^{2})

h (x) = ln (x + x^{2} - 11)

x, \frac{1}{2} x

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = x^{4} e^{x}

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