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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

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

= \frac{1}{ln ( 2 )} (\frac{1}{2} u tan^{2} (u) - \int \frac{1}{2} tan^{2} (u) d u)

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

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

Substitute back

u = 2^{x}

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

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

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

Add a constant to the solution

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

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