What is the integral of sec^8(4x)tan(4x) ?

The integral of sec^8(4x)tan(4x) is 1/32 sec^8(4x)+C

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integral of sec^8(4x)tan(4x)

\int sec^{8} (4 x) tan (4 x) d x

\frac{1}{32} sec^{8} (4 x) + C

Solution steps

\int sec^{8} (4 x) tan (4 x) d x

Apply u-substitution

: \int sec^{8} (u) tan (u) \frac{1}{4} d u

= \int sec^{8} (u) tan (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int sec^{8} (u) tan (u) d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{sin ( u )}{cos ^{9} ( u )} d u

Apply u-substitution

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

= \frac{1}{4} \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}{4} (- \int \frac{1}{v ^{9}} d v)

Apply the Power Rule

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

= \frac{1}{4} (- (- \frac{1}{8 v ^{8}}))

Substitute back

= \frac{1}{4} (- (- \frac{1}{8 cos ^{8} ( 4 x )}))

Simplify

\frac{1}{4} (- (- \frac{1}{8 cos ^{8} ( 4 x )})) : \frac{1}{32} sec^{8} (4 x)

= \frac{1}{32} sec^{8} (4 x)

Add a constant to the solution

= \frac{1}{32} sec^{8} (4 x) + C

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What is the integral of sec^8(4x)tan(4x) ?
The integral of sec^8(4x)tan(4x) is 1/32 sec^8(4x)+C