What is the integral of 1/(cos^4(x)) ?

The integral of 1/(cos^4(x)) is tan(x)+(tan^3(x))/3+C

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integral of 1/(cos^4(x))

\int \frac{1}{cos ^{4} ( x )} d x

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

Solution steps

\int \frac{1}{cos ^{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

ln (2 x), y = 0, y = 2

\int \frac{1}{x ^{2} + 8 x + 41} d x

\int 7 sec^{- 2} (x t a) n^{3} x

\frac{d}{d x} ((4 - x^{2})^{2})

x \to 2 lim (x^{2} - 3 x + 6)

What is the integral of 1/(cos^4(x)) ?
The integral of 1/(cos^4(x)) is tan(x)+(tan^3(x))/3+C