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

The integral of tan^2(2x)sec^3(2x) is 1/16 (-sec(2x)tan(2x)-ln|tan(2x)+sec(2x)|+2sec^4(2x)sin(2x))+C

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

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

\frac{1}{16} (- sec (2 x) tan (2 x) - ln ∣ tan (2 x) + sec (2 x) ∣ + 2 sec^{4} (2 x) sin (2 x)) + C

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Expand

(- 1 + sec^{2} (u)) sec^{3} (u) : - sec^{3} (u) + sec^{5} (u)

= \frac{1}{2} \cdot \int - sec^{3} (u) + sec^{5} (u) 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

= \frac{1}{2} (- \int sec^{3} (u) d u + \int sec^{5} (u) d u)

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

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= \frac{1}{2} (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

Substitute back

u = 2 x

= \frac{1}{2} (- (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣) + \frac{sec ^{4} ( 2 x ) sin ( 2 x )}{4} + \frac{3}{4} (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣))

Simplify

\frac{1}{2} (- (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣) + \frac{sec ^{4} ( 2 x ) sin ( 2 x )}{4} + \frac{3}{4} (\frac{1}{2} sec (2 x) tan (2 x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣)) : \frac{1}{16} (- sec (2 x) tan (2 x) - ln ∣ tan (2 x) + sec (2 x) ∣ + 2 sec^{4} (2 x) sin (2 x))

= \frac{1}{16} (- sec (2 x) tan (2 x) - ln ∣ tan (2 x) + sec (2 x) ∣ + 2 sec^{4} (2 x) sin (2 x))

Add a constant to the solution

= \frac{1}{16} (- sec (2 x) tan (2 x) - ln ∣ tan (2 x) + sec (2 x) ∣ + 2 sec^{4} (2 x) sin (2 x)) + C

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What is the integral of tan^2(2x)sec^3(2x) ?
The integral of tan^2(2x)sec^3(2x) is 1/16 (-sec(2x)tan(2x)-ln|tan(2x)+sec(2x)|+2sec^4(2x)sin(2x))+C