What is the integral of 7^xtan(7^x) ?

The integral of 7^xtan(7^x) is -1/(ln(7))ln|cos(7^x)|+C

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integral of 7^xtan(7^x)

\int 7^{x} tan (7^{x}) d x

- \frac{1}{ln ( 7 )} ln ∣ cos (7^{x}) ∣ + C

Solution steps

\int 7^{x} tan (7^{x}) d x

Apply u-substitution

: \int \frac{tan ( u )}{ln ( 7 )} d u

= \int \frac{tan ( u )}{ln ( 7 )} d u

Take the constant out:

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

= \frac{1}{ln ( 7 )} \cdot \int tan (u) d u

Rewrite using trig identities

= \frac{1}{ln ( 7 )} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= \frac{1}{ln ( 7 )} \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 ( 7 )} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{ln ( 7 )} (- ln ∣ v ∣)

Substitute back

= \frac{1}{ln ( 7 )} (- ln ∣ cos (7^{x}) ∣)

Simplify

= - \frac{1}{ln ( 7 )} ln ∣ cos (7^{x}) ∣

Add a constant to the solution

= - \frac{1}{ln ( 7 )} ln ∣ cos (7^{x}) ∣ + C

x \to \frac{π}{2} lim (e^{s i n (x)})

x \to 0 lim (\frac{x cos ( x )}{x})

\frac{d}{d x} (\frac{3}{x ^{2} + 9})

\frac{\partial}{\partial y} (y e^{- x^{2} + y})

\int \frac{x ^{2} + 3}{x x ^{2} - 9} d x

What is the integral of 7^xtan(7^x) ?
The integral of 7^xtan(7^x) is -1/(ln(7))ln|cos(7^x)|+C