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

The integral of sec^9(x)tan(x) is (sec^9(x))/9+C

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

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

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

Solution steps

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

Apply u-substitution

: \int u^{8} d u

= \int u^{8} d u

Apply the Power Rule

: \frac{u ^{9}}{9}

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

Substitute back

u = sec (x)

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

Add a constant to the solution

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

\frac{d}{d x} (sin (π \cdot x))

\frac{d}{d x} (\frac{7}{x + 8})

\frac{d}{d x} (A e^{- 3 x})

\int \frac{1}{ln ^{2} ( x )} d x

2 x^{2} + 3, (- 1, 2)

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