What is the integral of sec^{10}(x) ?

The integral of sec^{10}(x) is tan(x)+(4tan^3(x))/3+(6tan^5(x))/5+(4tan^7(x))/7+(tan^9(x))/9+C

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integral of sec^{10}(x)

\int sec^{10} (x) d x

tan (x) + \frac{4 tan ^{3} ( x )}{3} + \frac{6 tan ^{5} ( x )}{5} + \frac{4 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9} + C

Solution steps

\int sec^{10} (x) d x

Rewrite using trig identities

= \int sec^{2} (x) (1 + tan^{2} (x))^{4} d x

Apply u-substitution

: \int (1 + u^{2})^{4} d u

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

Expand

(1 + u^{2})^{4} : 1 + 4 u^{2} + 6 u^{4} + 4 u^{6} + u^{8}

= \int 1 + 4 u^{2} + 6 u^{4} + 4 u^{6} + u^{8} 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 4 u^{2} d u + \int 6 u^{4} d u + \int 4 u^{6} d u + \int u^{8} d u

\int 1 d u = u

\int 4 u^{2} d u = \frac{4 u ^{3}}{3}

\int 6 u^{4} d u = \frac{6 u ^{5}}{5}

\int 4 u^{6} d u = \frac{4 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= u + \frac{4 u ^{3}}{3} + \frac{6 u ^{5}}{5} + \frac{4 u ^{7}}{7} + \frac{u ^{9}}{9}

Substitute back

u = tan (x)

= tan (x) + \frac{4 tan ^{3} ( x )}{3} + \frac{6 tan ^{5} ( x )}{5} + \frac{4 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9}

Add a constant to the solution

= tan (x) + \frac{4 tan ^{3} ( x )}{3} + \frac{6 tan ^{5} ( x )}{5} + \frac{4 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9} + C

\int \frac{1}{1 + 2 x} d x

\frac{\partial}{\partial x} (y^{x^{7}})

f (t) = - 16 t^{2} + 32 t + 240, at t = 0

x \to + 2 lim (\frac{x ^{2} - 4}{x - 2})

x^{2}, 4 x

What is the integral of sec^{10}(x) ?
The integral of sec^{10}(x) is tan(x)+(4tan^3(x))/3+(6tan^5(x))/5+(4tan^7(x))/7+(tan^9(x))/9+C