What is the integral of (tan(x))^3sec(x) ?

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

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integral of (tan(x))^3sec(x)

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

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

Solution steps

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

Simplify

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

Rewrite using trig identities

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

Apply u-substitution

: \int - 1 + u^{2} d u

= \int - 1 + u^{2} d u

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 u + \int u^{2} d u

\int 1 d u = u

\int u^{2} d u = \frac{u ^{3}}{3}

= - u + \frac{u ^{3}}{3}

Substitute back

u = sec (x)

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

Add a constant to the solution

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

\int 84 d x

3^{s i n (3 x)}

\frac{\partial}{\partial x} (7 ln (\frac{2 x y}{5 z}))

\frac{\partial ^{2}}{\partial x ^{2}} (\frac{x}{x ^{2} + y ^{2}})

\int 2 x e^{7 x} d x

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