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

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

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integral of tan^3(2x)

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

- \frac{1}{2} ln ∣ sec (2 x) ∣ + \frac{sec ^{2} ( 2 x )}{4} + C

Solution steps

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

Rewrite using trig identities

= \int (- 1 + sec^{2} (2 x)) tan (2 x) d x

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{2 u} d u

= \int \frac{- 1 + u ^{2}}{2 u} d u

Expand

\frac{- 1 + u ^{2}}{2 u} : - \frac{1}{2 u} + \frac{u}{2}

= \int - \frac{1}{2 u} + \frac{u}{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{2 u} d u + \int \frac{u}{2} d u

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{u}{2} d u = \frac{u ^{2}}{4}

= - \frac{1}{2} ln ∣ u ∣ + \frac{u ^{2}}{4}

Substitute back

u = sec (2 x)

= - \frac{1}{2} ln ∣ sec (2 x) ∣ + \frac{sec ^{2} ( 2 x )}{4}

Add a constant to the solution

= - \frac{1}{2} ln ∣ sec (2 x) ∣ + \frac{sec ^{2} ( 2 x )}{4} + C

y = f (x), at x = 7

5^{7 - x^{2}}

8 t

x^{2} - 6 x, - x^{2} + 4 x

f (t) = 3 t^{2} - t

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