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

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

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

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

\frac{1}{4} (- x + tan (x)) + C

Solution steps

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

Take the constant out:

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

= \frac{1}{4} \cdot \int tan^{2} (x) d x

Rewrite using trig identities

= \frac{1}{4} \cdot \int - 1 + sec^{2} (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

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

\int 1 d x = x

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

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

Add a constant to the solution

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

x^{3 x}

\frac{d}{d x} (\frac{8 x}{5 - tan ( x )})

y^{^{'}} - \frac{6}{x} y = x^{- 4} y^{5}, y (1) = 1

\int_{0}^{\infty} 4 e^{- 4 x} d x

\int \frac{x}{x - 2} d x

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