What is the integral of tan^3(x)sec^9(x) ?

The integral of tan^3(x)sec^9(x) is (sec^{11}(x)}{11}-\frac{sec^9(x))/9+C

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integral of tan^3(x)sec^9(x)

\int tan^{3} (x) sec^{9} (x) d x

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

Solution steps

\int tan^{3} (x) sec^{9} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \int 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^{10} d u - \int u^{8} d u

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

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

= \frac{u ^{11}}{11} - \frac{u ^{9}}{9}

Substitute back

u = sec (x)

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

Add a constant to the solution

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

\int x^{2} (x + 1)^{3} d x

(3 y^{2} + 2 x y) - (2 x y + x^{2}) \frac{d y}{d x} = 0

3 (s^{3} - s^{2})

f (r) = \frac{6}{r ^{3}}

x \to 2 π lim (x \cdot csc (x))

What is the integral of tan^3(x)sec^9(x) ?
The integral of tan^3(x)sec^9(x) is (sec^{11}(x)}{11}-\frac{sec^9(x))/9+C