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

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

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

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

\frac{1}{2} e^{t a n (2 x)} + C

Solution steps

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

Apply u-substitution

= \int e^{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

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

Apply u-substitution

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

Use the common integral:

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

= \frac{1}{2} e^{v}

Substitute back

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

Add a constant to the solution

= \frac{1}{2} e^{t a n (2 x)} + C

\int \frac{1}{3 x ^{4}} d x

\int_{- 1}^{1} \frac{1}{2} (4 - 4 x^{2})^{2} d x

\frac{\partial}{\partial t} (\frac{s - 3 r}{t})

y^{^{''}} - y = x

\frac{d}{d x} (\frac{7}{3 x ^{4}})

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