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

The integral of 2^xtan(3-2^x) is 1/(ln(2))ln|cos(3-2^x)|+C

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integral of 2^xtan(3-2^x)

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

\frac{1}{ln ( 2 )} ln ∣ cos (3 - 2^{x}) ∣ + C

Solution steps

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

Apply u-substitution

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

= \int - \frac{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 tan (u) d u

Rewrite using trig identities

= - \frac{1}{ln ( 2 )} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= - \frac{1}{ln ( 2 )} \cdot \int - \frac{1}{v} d v

Take the constant out:

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

= - \frac{1}{ln ( 2 )} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{ln ( 2 )} (- ln ∣ v ∣)

Substitute back

= - \frac{1}{ln ( 2 )} (- ln ∣ cos (3 - 2^{x}) ∣)

Simplify

= \frac{1}{ln ( 2 )} ln ∣ cos (3 - 2^{x}) ∣

Add a constant to the solution

= \frac{1}{ln ( 2 )} ln ∣ cos (3 - 2^{x}) ∣ + C

2 x y \frac{d y}{d x} = 5

x \to 2 lim (3 x + 4)

f (t) = sin (t)

\frac{d}{d x} (f (x) (sin (x)))

\int \frac{1}{9 + 16 x ^{2}} d x

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