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

The integral of tan^9(x) is ln|sec(x)|-2sec^2(x)+(3sec^4(x))/2-(2sec^6(x))/3+(sec^8(x))/8+C

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integral of tan^9(x)

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

ln ∣ sec (x) ∣ - 2 sec^{2} (x) + \frac{3 sec ^{4} ( x )}{2} - \frac{2 sec ^{6} ( x )}{3} + \frac{sec ^{8} ( x )}{8} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{( - 1 + u ^{2} ) ^{4}}{u} : \frac{1}{u} - 4 u + 6 u^{3} - 4 u^{5} + u^{7}

= \int \frac{1}{u} - 4 u + 6 u^{3} - 4 u^{5} + u^{7} 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 \frac{1}{u} d u - \int 4 u d u + \int 6 u^{3} d u - \int 4 u^{5} d u + \int u^{7} d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int 6 u^{3} d u = \frac{3 u ^{4}}{2}

\int 4 u^{5} d u = \frac{2 u ^{6}}{3}

\int u^{7} d u = \frac{u ^{8}}{8}

= ln ∣ u ∣ - 2 u^{2} + \frac{3 u ^{4}}{2} - \frac{2 u ^{6}}{3} + \frac{u ^{8}}{8}

Substitute back

u = sec (x)

= ln ∣ sec (x) ∣ - 2 sec^{2} (x) + \frac{3 sec ^{4} ( x )}{2} - \frac{2 sec ^{6} ( x )}{3} + \frac{sec ^{8} ( x )}{8}

Add a constant to the solution

= ln ∣ sec (x) ∣ - 2 sec^{2} (x) + \frac{3 sec ^{4} ( x )}{2} - \frac{2 sec ^{6} ( x )}{3} + \frac{sec ^{8} ( x )}{8} + C

\int e^{- x^{2} - y^{2}}

f (x) = \frac{x}{x ^{2} + 8}

f (x) = e^{x^{3} l n (x)}

\int \frac{9 sin ( x ) + 4 sec ( x )}{tan ( x )} d x

f (x) = \frac{6}{3 x + 1}

What is the integral of tan^9(x) ?
The integral of tan^9(x) is ln|sec(x)|-2sec^2(x)+(3sec^4(x))/2-(2sec^6(x))/3+(sec^8(x))/8+C