What is the integral of e^{1/(sec(2x))}sin(2x) ?

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

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integral of e^{1/(sec(2x))}sin(2x)

\int e^{\frac{1}{s e c ( 2 x )}} sin (2 x) d x

- \frac{1}{2} e^{c o s (2 x)} + C

Solution steps

\int e^{\frac{1}{s e c ( 2 x )}} sin (2 x) d x

Use the basic trigonometric identity:

\frac{1}{sec ( x )} = cos (x)

= \int e^{c o s (2 x)} sin (2 x) d x

Apply u-substitution

: \int e^{c o s (u)} sin (u) \frac{1}{2} d u

= \int e^{c o s (u)} sin (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^{c o s (u)} sin (u) d u

Apply u-substitution

: \int - e^{v} d v

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

Take the constant out:

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

= \frac{1}{2} (- \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^{c o s (2 x)})

Simplify

= - \frac{1}{2} e^{c o s (2 x)}

Add a constant to the solution

= - \frac{1}{2} e^{c o s (2 x)} + C

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