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

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

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integral of sec^8(x)tan(x)

\int sec^{8} (x) tan (x) d x

\frac{1}{8} sec^{8} (x) + C

Solution steps

\int sec^{8} (x) tan (x) d x

Rewrite using trig identities

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

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 (x)

= - (- \frac{1}{8 cos ^{8} ( x )})

Simplify

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

= \frac{1}{8} sec^{8} (x)

Add a constant to the solution

= \frac{1}{8} sec^{8} (x) + C

\frac{d}{d x} (- x + 2)

f (t) = cos (t) + t^{2} sin (t)

\int - 9 e^{- x} d x

\frac{d}{d x} (\frac{25}{x ^{2}} - \frac{5}{x})

y^{^{''}} - 2 y^{^{'}} + y = e^{3 x}

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