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 (x) = \frac{cos ( x )}{1 + sin ( x )}

\int \frac{6 x ^{5} + x ^{\frac{1}{4}} - 3}{- 5 x ^{\frac{3}{2}}} d x

\frac{d y}{d x} = 0.001 y (100 - y)

\frac{\partial}{\partial x} (x^{2} y + y^{3})

- x^{2} + 1.907, 0, 0.266

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