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

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

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integral of sec^9(θ)tan(θ)

\int sec^{9} (θ) tan (θ) d θ

\frac{sec ^{9} ( θ )}{9} + C

Solution steps

\int sec^{9} (θ) tan (θ) d θ

Apply u-substitution

= \int u^{8} d u

Apply the Power Rule

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

Substitute back

u = sec (θ)

= \frac{sec ^{9} ( θ )}{9}

Add a constant to the solution

= \frac{sec ^{9} ( θ )}{9} + C

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