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

The integral of tan^3(2x)sec^2(2x) is 1/8 tan^4(2x)+C

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integral of tan^3(2x)sec^2(2x)

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

\frac{1}{8} tan^{4} (2 x) + C

Solution steps

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

Apply u-substitution

= \int tan^{3} (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

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

Rewrite using trig identities

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

Apply u-substitution

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

Apply the Power Rule

= \frac{1}{2} \cdot \frac{v ^{4}}{4}

Substitute back

= \frac{1}{2} \cdot \frac{tan ^{4} ( 2 x )}{4}

Simplify

\frac{1}{2} \cdot \frac{tan ^{4} ( 2 x )}{4} : \frac{1}{8} tan^{4} (2 x)

= \frac{1}{8} tan^{4} (2 x)

Add a constant to the solution

= \frac{1}{8} tan^{4} (2 x) + C

What is the integral of tan^3(2x)sec^2(2x) ?
The integral of tan^3(2x)sec^2(2x) is 1/8 tan^4(2x)+C