What is the integral of 5(tan^2(x)+tan^4(x)) ?

The integral of 5(tan^2(x)+tan^4(x)) is 5/3 tan^3(x)+C

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integral of 5(tan^2(x)+tan^4(x))

\int 5 (tan^{2} (x) + tan^{4} (x)) d x

\frac{5}{3} tan^{3} (x) + C

Solution steps

\int 5 (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

= 5 \cdot \int tan^{2} (x) + tan^{4} (x) d x

Rewrite using trig identities

= 5 \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

= 5 (- \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)

= 5 (- x + tan (x) + \frac{1}{3} tan^{3} (x) + x - tan (x))

Simplify

5 (- x + tan (x) + \frac{1}{3} tan^{3} (x) + x - tan (x)) : \frac{5}{3} tan^{3} (x)

= \frac{5}{3} tan^{3} (x)

Add a constant to the solution

= \frac{5}{3} tan^{3} (x) + C

\int (e^{x} + e^{- x}) d x

\int \frac{x ^{3}}{1 - x ^{4}} d x

x = - 5, x = 1, y = 11 x, y = x^{2} - 12

2 x^{4}

\frac{s + 4}{s ^{2} + 6 s + 25}

What is the integral of 5(tan^2(x)+tan^4(x)) ?
The integral of 5(tan^2(x)+tan^4(x)) is 5/3 tan^3(x)+C