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

The integral of 2tan^2(2x)sec^2(2x) is 1/3 tan^3(2x)+C

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

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

\frac{1}{3} tan^{3} (2 x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int v^{2} d v

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

Apply the Power Rule

: \frac{v ^{3}}{3}

= 2 \cdot \frac{1}{2} \cdot \frac{v ^{3}}{3}

Substitute back

= 2 \cdot \frac{1}{2} \cdot \frac{tan ^{3} ( 2 x )}{3}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{tan ^{3} ( 2 x )}{3} : \frac{1}{3} tan^{3} (2 x)

= \frac{1}{3} tan^{3} (2 x)

Add a constant to the solution

= \frac{1}{3} tan^{3} (2 x) + C

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What is the integral of 2tan^2(2x)sec^2(2x) ?
The integral of 2tan^2(2x)sec^2(2x) is 1/3 tan^3(2x)+C