What is the integral of sec^8(2x)tan(2x) ?

The integral of sec^8(2x)tan(2x) is 1/16 sec^8(2x)+C

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integral of sec^8(2x)tan(2x)

\int sec^{8} (2 x) tan (2 x) d x

\frac{1}{16} sec^{8} (2 x) + C

Solution steps

\int sec^{8} (2 x) tan (2 x) d x

Apply u-substitution

= \int sec^{8} (u) tan (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sec^{8} (u) tan (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{sin ( u )}{cos ^{9} ( u )} d u

Apply u-substitution

= \frac{1}{2} \cdot \int - \frac{1}{v ^{9}} d v

Take the constant out:

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

= \frac{1}{2} (- \int \frac{1}{v ^{9}} d v)

Apply the Power Rule

= \frac{1}{2} (- (- \frac{1}{8 v ^{8}}))

Substitute back

= \frac{1}{2} (- (- \frac{1}{8 cos ^{8} ( 2 x )}))

Simplify

\frac{1}{2} (- (- \frac{1}{8 cos ^{8} ( 2 x )})) : \frac{1}{16} sec^{8} (2 x)

= \frac{1}{16} sec^{8} (2 x)

Add a constant to the solution

= \frac{1}{16} sec^{8} (2 x) + C

s l o p e (2.4) (6.6)

\frac{d}{d x} (2 x^{3} y)

\int \frac{1}{x - 3} d x

a re a y = 12 - x, y = x, y = 1

d er i v a t i v e e^{t} cos (6 t)

What is the integral of sec^8(2x)tan(2x) ?
The integral of sec^8(2x)tan(2x) is 1/16 sec^8(2x)+C