What is the integral of-12sec^2(-4x)tan^3(-4x) ?

The integral of-12sec^2(-4x)tan^3(-4x) is 3/4 tan^4(4x)+C

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integral of-12sec^2(-4x)tan^3(-4x)

\int - 12 sec^{2} (- 4 x) tan^{3} (- 4 x) d x

\frac{3}{4} tan^{4} (4 x) + C

Solution steps

\int - 12 sec^{2} (- 4 x) tan^{3} (- 4 x) d x

Take the constant out:

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

= - 12 \cdot \int sec^{2} (- 4 x) tan^{3} (- 4 x) d x

Simplify

sec^{2} (- 4 x) tan^{3} (- 4 x) : - tan^{3} (4 x) sec^{2} (4 x)

= - 12 \cdot \int - tan^{3} (4 x) sec^{2} (4 x) d x

Take the constant out:

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

= - 12 (- \int tan^{3} (4 x) sec^{2} (4 x) d x)

Apply u-substitution

: \int tan^{3} (u) sec^{2} (u) \frac{1}{4} d u

= - 12 (- \int tan^{3} (u) sec^{2} (u) \frac{1}{4} d u)

Take the constant out:

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

= - 12 (- \frac{1}{4} \cdot \int tan^{3} (u) sec^{2} (u) d u)

Rewrite using trig identities

= - 12 (- \frac{1}{4} \cdot \int (1 + tan^{2} (u)) tan^{3} (u) d u)

Apply u-substitution

: \int v^{3} d v

= - 12 (- \frac{1}{4} \cdot \int v^{3} d v)

Apply the Power Rule

: \frac{v ^{4}}{4}

= - 12 (- \frac{1}{4} \cdot \frac{v ^{4}}{4})

Substitute back

= - 12 (- \frac{1}{4} \cdot \frac{tan ^{4} ( 4 x )}{4})

Simplify

- 12 (- \frac{1}{4} \cdot \frac{tan ^{4} ( 4 x )}{4}) : \frac{3}{4} tan^{4} (4 x)

= \frac{3}{4} tan^{4} (4 x)

Add a constant to the solution

= \frac{3}{4} tan^{4} (4 x) + C

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What is the integral of-12sec^2(-4x)tan^3(-4x) ?
The integral of-12sec^2(-4x)tan^3(-4x) is 3/4 tan^4(4x)+C