What is the integral of e^{2x}tan^2(e^{2x}) ?

The integral of e^{2x}tan^2(e^{2x}) is 1/2 (-e^{2x}+tan(e^{2x}))+C

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integral of e^{2x}tan^2(e^{2x})

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int tan^{2} (v) d v

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

Rewrite using trig identities

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

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 1 d v + \int sec^{2} (v) d v)

\int 1 d v = v

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

= \frac{1}{2} (- v + tan (v))

Substitute back

= \frac{1}{2} (- e^{2 x} + tan (e^{2 x}))

Add a constant to the solution

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

\frac{d}{d x} (2 a cos (2 x) - 2 b sin (2 x))

r \frac{2 x}{3 x ^{2} - 4}

f (x) = \frac{x ^{2} - 2 x}{x}

(x^{2} - 2 x)^{3}

in (x^{2})

What is the integral of e^{2x}tan^2(e^{2x}) ?
The integral of e^{2x}tan^2(e^{2x}) is 1/2 (-e^{2x}+tan(e^{2x}))+C