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

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

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

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

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

Solution steps

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

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