What is the integral of tan^5(x)sec^9(x) ?

The integral of tan^5(x)sec^9(x) is (sec^{13}(x)}{13}-(2sec^{11}(x))/(11)+\frac{sec^9(x))/9+C

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integral of tan^5(x)sec^9(x)

\int tan^{5} (x) sec^{9} (x) d x

\frac{sec ^{13} ( x )}{13} - \frac{2 sec ^{11} ( x )}{11} + \frac{sec ^{9} ( x )}{9} + C

Solution steps

\int tan^{5} (x) sec^{9} (x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (x))^{2} tan (x) sec^{9} (x) d x

Apply u-substitution

: \int u^{8} (u^{2} - 1)^{2} d u

= \int u^{8} (u^{2} - 1)^{2} d u

Expand

u^{8} (u^{2} - 1)^{2} : u^{12} - 2 u^{10} + u^{8}

= \int u^{12} - 2 u^{10} + 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 u^{12} d u - \int 2 u^{10} d u + \int u^{8} d u

\int u^{12} d u = \frac{u ^{13}}{13}

\int 2 u^{10} d u = \frac{2 u ^{11}}{11}

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

= \frac{u ^{13}}{13} - \frac{2 u ^{11}}{11} + \frac{u ^{9}}{9}

Substitute back

u = sec (x)

= \frac{sec ^{13} ( x )}{13} - \frac{2 sec ^{11} ( x )}{11} + \frac{sec ^{9} ( x )}{9}

Add a constant to the solution

= \frac{sec ^{13} ( x )}{13} - \frac{2 sec ^{11} ( x )}{11} + \frac{sec ^{9} ( x )}{9} + C

\int \frac{8 x ^{2} + 9 x + 8}{( x ^{2} + 1 ) ^{2}} d x

n = 0 \sum \infty (\frac{3}{7})^{n}

\frac{d y}{d x} = 9 x^{3}

\frac{d}{d x} (cot (π) x \cdot sec (π) x)

5 x + 6

What is the integral of tan^5(x)sec^9(x) ?
The integral of tan^5(x)sec^9(x) is (sec^{13}(x)}{13}-(2sec^{11}(x))/(11)+\frac{sec^9(x))/9+C