What is the integral of 14tan^3(x)sec(x) ?

The integral of 14tan^3(x)sec(x) is 14(-sec(x)+(sec^3(x))/3)+C

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integral of 14tan^3(x)sec(x)

\int 14 tan^{3} (x) sec (x) d x

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

Solution steps

\int 14 tan^{3} (x) sec (x) d x

Take the constant out:

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

= 14 \cdot \int tan^{3} (x) sec (x) d x

Rewrite using trig identities

= 14 \cdot \int (- 1 + sec^{2} (x)) tan (x) sec (x) d x

Apply u-substitution

= 14 \cdot \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

= 14 (- \int 1 d u + \int u^{2} d u)

\int 1 d u = u

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

= 14 (- u + \frac{u ^{3}}{3})

Substitute back

u = sec (x)

= 14 (- sec (x) + \frac{sec ^{3} ( x )}{3})

Add a constant to the solution

= 14 (- sec (x) + \frac{sec ^{3} ( x )}{3}) + C

x \to 0 + lim (x^{x^{\frac{1}{2}}})

\int x sin (3 x) d x

a re a y = 2 cos (x), y = 6 - 4 cos (x), 0 \leq x \leq 2 π

\frac{\partial}{\partial x} (4 - 2 x^{2} - y^{2})

\frac{d}{d x} (e^{- b x^{2}})

What is the integral of 14tan^3(x)sec(x) ?
The integral of 14tan^3(x)sec(x) is 14(-sec(x)+(sec^3(x))/3)+C