What is the integral of e^{tan(4x)}sec^2(4x) ?

The integral of e^{tan(4x)}sec^2(4x) is 1/4 e^{tan(4x)}+C

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integral of e^{tan(4x)}sec^2(4x)

\int e^{t a n (4 x)} sec^{2} (4 x) d x

\frac{1}{4} e^{t a n (4 x)} + C

Solution steps

\int e^{t a n (4 x)} sec^{2} (4 x) d x

Apply u-substitution

: \int e^{t a n (u)} sec^{2} (u) \frac{1}{4} d u

= \int e^{t a n (u)} sec^{2} (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{t a n (u)} sec^{2} (u) d u

Apply u-substitution

: \int e^{v} d v

= \frac{1}{4} \cdot \int e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{4} e^{v}

Substitute back

= \frac{1}{4} e^{t a n (4 x)}

Add a constant to the solution

= \frac{1}{4} e^{t a n (4 x)} + C

\frac{1}{1 - x}

\int_{e}^{e^{2}} \frac{1}{x ln ( x )} d x

f (x) = \frac{e ^{x}}{sin ( 2 x )}

\frac{d}{d x} (\frac{40}{1 + 79 e ^{- x}})

\frac{d}{d x} (\frac{x}{x - 7})

What is the integral of e^{tan(4x)}sec^2(4x) ?
The integral of e^{tan(4x)}sec^2(4x) is 1/4 e^{tan(4x)}+C