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

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

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

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

\frac{1}{2} (sec (2 x) + ln ∣ tan (x) ∣) + C

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

\frac{cos ( x )}{sin ( x )} = cot (x)

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

Rewrite using trig identities

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

Apply Integration By Parts

: sec (u) - \int - cot (u) csc (u) tan (u) d u

= \frac{1}{2} (sec (u) - \int - cot (u) csc (u) tan (u) d u)

\int - cot (u) csc (u) tan (u) d u = - ln tan (\frac{u}{2})

= \frac{1}{2} (sec (u) - (- ln tan (\frac{u}{2})))

Substitute back

u = 2 x

= \frac{1}{2} (sec (2 x) - (- ln tan (\frac{2 x}{2})))

Simplify

\frac{1}{2} (sec (2 x) - (- ln tan (\frac{2 x}{2}))) : \frac{1}{2} (sec (2 x) + ln ∣ tan (x) ∣)

= \frac{1}{2} (sec (2 x) + ln ∣ tan (x) ∣)

Add a constant to the solution

= \frac{1}{2} (sec (2 x) + ln ∣ tan (x) ∣) + C

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