What is the integral of 13tan^3(x)sec(x) ?

The integral of 13tan^3(x)sec(x) is 13(-sec(x)+(sec^3(x))/3)+C

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integral of 13tan^3(x)sec(x)

\int 13 tan^{3} (x) sec (x) d x

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

Solution steps

\int 13 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

= 13 \cdot \int tan^{3} (x) sec (x) d x

Rewrite using trig identities

= 13 \cdot \int (- 1 + sec^{2} (x)) tan (x) sec (x) d x

Apply u-substitution

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

= 13 \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

= 13 (- \int 1 d u + \int u^{2} d u)

\int 1 d u = u

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

= 13 (- u + \frac{u ^{3}}{3})

Substitute back

u = sec (x)

= 13 (- sec (x) + \frac{sec ^{3} ( x )}{3})

Add a constant to the solution

= 13 (- sec (x) + \frac{sec ^{3} ( x )}{3}) + C

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

\frac{d y}{d t} + \frac{3 t}{t ^{2} + 1} y = \frac{6 t}{t ^{2} + 1}

\int - 3 x^{4} d x

\frac{d}{d x} (\frac{2 x - 1}{x + 9})

(x - 1)^{3}, x - 1, 0, 1

What is the integral of 13tan^3(x)sec(x) ?
The integral of 13tan^3(x)sec(x) is 13(-sec(x)+(sec^3(x))/3)+C