What is the integral of tan^2(x)(sec^2(x)-1) ?

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

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integral of tan^2(x)(sec^2(x)-1)

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

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

Solution steps

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

Expand

tan^{2} (x) (sec^{2} (x) - 1) : tan^{2} (x) sec^{2} (x) - tan^{2} (x)

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} ((x^{3} + x)^{\frac{3}{2}})

\frac{d}{d x} (2^{c o s (x)})

n = 0 \sum \infty 2^{3 - 2 n}

\int \frac{x}{( 1 + 2 x ) ^{2}} d x

\frac{\partial}{\partial t} (2 s^{2} t^{2})

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