What is the integral of sec^8(θ)tan(θ) ?

The integral of sec^8(θ)tan(θ) is 1/8 sec^8(θ)+C

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integral of sec^8(θ)tan(θ)

\int sec^{8} (θ) tan (θ) d θ

\frac{1}{8} sec^{8} (θ) + C

Solution steps

\int sec^{8} (θ) tan (θ) d θ

Rewrite using trig identities

= \int \frac{sin ( θ )}{cos ^{9} ( θ )} d θ

Apply u-substitution

: \int - \frac{1}{u ^{9}} d u

= \int - \frac{1}{u ^{9}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u ^{9}} d u

Apply the Power Rule

: - \frac{1}{8 u ^{8}}

= - (- \frac{1}{8 u ^{8}})

Substitute back

u = cos (θ)

= - (- \frac{1}{8 cos ^{8} ( θ )})

Simplify

- (- \frac{1}{8 cos ^{8} ( θ )}) : \frac{1}{8} sec^{8} (θ)

= \frac{1}{8} sec^{8} (θ)

Add a constant to the solution

= \frac{1}{8} sec^{8} (θ) + C

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What is the integral of sec^8(θ)tan(θ) ?
The integral of sec^8(θ)tan(θ) is 1/8 sec^8(θ)+C