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

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

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

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

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

Solution steps

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

x \to 3 lim (- x + 2)

\int_{2}^{5} \frac{1}{x ^{2} - 4} d x

\int \frac{x ^{5}}{( x ^{2} + 4 ) ^{2}} d x

\frac{d}{d x} (6 x^{3} - 4 x^{2} + 5 x)

f (x) = \frac{9 x}{x + 4}, at x = 8

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