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

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

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

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

\frac{2}{3} tan^{3} (x) + C

Solution steps

\int 2 (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

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

Rewrite using trig identities

= 2 \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

= 2 (- \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)

= 2 (- x + tan (x) + \frac{1}{3} tan^{3} (x) + x - tan (x))

Simplify

2 (- x + tan (x) + \frac{1}{3} tan^{3} (x) + x - tan (x)) : \frac{2}{3} tan^{3} (x)

= \frac{2}{3} tan^{3} (x)

Add a constant to the solution

= \frac{2}{3} tan^{3} (x) + C

\int x arccot (x) d x

f^{^{''}} (x) = x^{- \frac{3}{2}}, f^{^{'}} (4) = 7, f (0) = 0

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

\frac{2 \cdot ( s - 1 ) \cdot e ^{- 2 s}}{s ^{2} - 2 s + 2}

y = x^{2} + 2, (- 4, 18)

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