What is the integral of sec^5(x)tan^5(x) ?

The integral of sec^5(x)tan^5(x) is (sec^9(x))/9-(2sec^7(x))/7+(sec^5(x))/5+C

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integral of sec^5(x)tan^5(x)

\int sec^{5} (x) tan^{5} (x) d x

\frac{sec ^{9} ( x )}{9} - \frac{2 sec ^{7} ( x )}{7} + \frac{sec ^{5} ( x )}{5} + C

Solution steps

\int sec^{5} (x) tan^{5} (x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (x))^{2} tan (x) sec^{5} (x) d x

Apply u-substitution

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

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

Expand

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

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

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

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

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

= \frac{u ^{9}}{9} - \frac{2 u ^{7}}{7} + \frac{u ^{5}}{5}

Substitute back

u = sec (x)

= \frac{sec ^{9} ( x )}{9} - \frac{2 sec ^{7} ( x )}{7} + \frac{sec ^{5} ( x )}{5}

Add a constant to the solution

= \frac{sec ^{9} ( x )}{9} - \frac{2 sec ^{7} ( x )}{7} + \frac{sec ^{5} ( x )}{5} + C

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What is the integral of sec^5(x)tan^5(x) ?
The integral of sec^5(x)tan^5(x) is (sec^9(x))/9-(2sec^7(x))/7+(sec^5(x))/5+C