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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= \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

= - \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)

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

Simplify

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

Add a constant to the solution

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

n = 1 \sum \infty (- 5)^{n} x^{n}

arcsin (x^{2})

\frac{d}{d x} (cos (\frac{x}{2}) \frac{1}{2})

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