What is the integral of sec^{13}(θ)tan(θ) ?

The integral of sec^{13}(θ)tan(θ) is (sec^{13}(θ))/(13)+C

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integral of sec^{13}(θ)tan(θ)

\int sec^{13} (θ) tan (θ) d θ

\frac{sec ^{13} ( θ )}{13} + C

Solution steps

\int sec^{13} (θ) tan (θ) d θ

Apply u-substitution

: \int u^{12} d u

= \int u^{12} d u

Apply the Power Rule

: \frac{u ^{13}}{13}

= \frac{u ^{13}}{13}

Substitute back

u = sec (θ)

= \frac{sec ^{13} ( θ )}{13}

Add a constant to the solution

= \frac{sec ^{13} ( θ )}{13} + C

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What is the integral of sec^{13}(θ)tan(θ) ?
The integral of sec^{13}(θ)tan(θ) is (sec^{13}(θ))/(13)+C