What is the integral of sec^4(x) ?

The integral of sec^4(x) is tan(x)+(tan^3(x))/3+C

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integral of sec^4(x)

\int sec^{4} (x) d x

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

Solution steps

\int sec^{4} (x) d x

Rewrite using trig identities

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

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

Add a constant to the solution

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

f (t) = t^{4}

(1436.53) \cdot (1.01395)^{t}

n = 0 \sum \infty 4 (\frac{2}{3})^{n - 1}

\int \frac{16 x ^{2} - 1}{x} d x

\frac{d}{d x} (x^{2} (x - 12)^{2})

What is the integral of sec^4(x) ?
The integral of sec^4(x) is tan(x)+(tan^3(x))/3+C