What is the integral of sec^8(t)anx ?

The integral of sec^8(t)anx is (anx^2sec^8(t))/2+C

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integral of sec^8(t)anx

\int sec^{8} (t) an x d x

\frac{an x ^{2} sec ^{8} ( t )}{2} + C

Solution steps

\int sec^{8} (t) an x d x

Take the constant out:

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

= sec^{8} (t) an \cdot \int x d x

Apply the Power Rule

: \frac{x ^{2}}{2}

= sec^{8} (t) an \frac{x ^{2}}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{an x ^{2} sec ^{8} ( t )}{2}

Add a constant to the solution

= \frac{an x ^{2} sec ^{8} ( t )}{2} + C

\int - \frac{1}{9} x^{- \frac{10}{9}} d x

\frac{d}{d x} (\frac{2}{x ^{13}})

\int (tan (x) - sin (x) sin (y)) d x

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What is the integral of sec^8(t)anx ?
The integral of sec^8(t)anx is (anx^2sec^8(t))/2+C