What is the integral of 1/(cos^6(x)) ?

The integral of 1/(cos^6(x)) is tan(x)+(2tan^3(x))/3+(tan^5(x))/5+C

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integral of 1/(cos^6(x))

\int \frac{1}{cos ^{6} ( x )} d x

tan (x) + \frac{2 tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5} + C

Solution steps

\int \frac{1}{cos ^{6} ( x )} d x

Rewrite using trig identities

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

Apply u-substitution

: \int (1 + u^{2})^{2} d u

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

Expand

(1 + u^{2})^{2} : 1 + 2 u^{2} + u^{4}

= \int 1 + 2 u^{2} + u^{4} 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 2 u^{2} d u + \int u^{4} d u

\int 1 d u = u

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

\int u^{4} d u = \frac{u ^{5}}{5}

= u + \frac{2 u ^{3}}{3} + \frac{u ^{5}}{5}

Substitute back

u = tan (x)

= tan (x) + \frac{2 tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5}

Add a constant to the solution

= tan (x) + \frac{2 tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5} + C

x \to \infty lim (\frac{( 5 + 2 e ^{- 2 x} )}{ln ( x )})

17 - x^{2}

x \to 1 lim (\frac{1}{x})

4 - x^{2}, x = 0

- 2 x^{2} + 8 x, [- 1, 2]

What is the integral of 1/(cos^6(x)) ?
The integral of 1/(cos^6(x)) is tan(x)+(2tan^3(x))/3+(tan^5(x))/5+C