What is the integral of 6e^{4tan(2x)}sec^2(2x) ?

The integral of 6e^{4tan(2x)}sec^2(2x) is 3/4 e^{4tan(2x)}+C

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integral of 6e^{4tan(2x)}sec^2(2x)

\int 6 e^{4 t a n (2 x)} sec^{2} (2 x) d x

\frac{3}{4} e^{4 t a n (2 x)} + C

Solution steps

\int 6 e^{4 t a n (2 x)} sec^{2} (2 x) d x

Take the constant out:

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

= 6 \cdot \int e^{4 t a n (2 x)} sec^{2} (2 x) d x

Apply u-substitution

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

= 6 \cdot \int e^{4 t a n (u)} sec^{2} (u) \frac{1}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int e^{4 t a n (u)} sec^{2} (u) d u

Apply u-substitution

: \int e^{4 v} d v

= 6 \cdot \frac{1}{2} \cdot \int e^{4 v} d v

Apply u-substitution

: \int e^{w} \frac{1}{4} d w

= 6 \cdot \frac{1}{2} \cdot \int e^{w} \frac{1}{4} d w

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int e^{w} d w

Use the common integral:

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

= 6 \cdot \frac{1}{2} \cdot \frac{1}{4} e^{w}

Substitute back

= 6 \cdot \frac{1}{2} \cdot \frac{1}{4} e^{4 t a n (2 x)}

Simplify

6 \cdot \frac{1}{2} \cdot \frac{1}{4} e^{4 t a n (2 x)} : \frac{3}{4} e^{4 t a n (2 x)}

= \frac{3}{4} e^{4 t a n (2 x)}

Add a constant to the solution

= \frac{3}{4} e^{4 t a n (2 x)} + C

\frac{d}{d x} (\frac{- x ^{2}}{y ^{2}})

\int \frac{x}{1 + 2 x} d x

\int (3 x zy + x y^{2} + 2 z x^{2}) d x

x \to 3 + lim (f (x))

f (x) = \frac{x ^{2}}{2} \cdot cos^{3} (\frac{1}{x})

What is the integral of 6e^{4tan(2x)}sec^2(2x) ?
The integral of 6e^{4tan(2x)}sec^2(2x) is 3/4 e^{4tan(2x)}+C