What is the integral of (3^{cos(2z)})/(csc(2z)) ?

The integral of (3^{cos(2z)})/(csc(2z)) is -(3^{cos(2z)})/(2ln(3))+C

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integral of (3^{cos(2z)})/(csc(2z))

\int \frac{3 ^{c o s (2 z)}}{csc ( 2 z )} d z

- \frac{3 ^{c o s (2 z)}}{2 ln ( 3 )} + C

Solution steps

\int \frac{3 ^{c o s (2 z)}}{csc ( 2 z )} d z

Use the basic trigonometric identity:

\frac{1}{csc ( x )} = sin (x)

= \int 3^{c o s (2 z)} sin (2 z) d z

Apply u-substitution

: \int 3^{c o s (u)} sin (u) \frac{1}{2} d u

= \int 3^{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 3^{c o s (u)} sin (u) d u

Apply u-substitution

: \int - 3^{v} d v

= \frac{1}{2} \cdot \int - 3^{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 3^{v} d v)

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{2} (- \frac{3 ^{v}}{ln ( 3 )})

Substitute back

= \frac{1}{2} (- \frac{3 ^{c o s (2 z)}}{ln ( 3 )})

Simplify

\frac{1}{2} (- \frac{3 ^{c o s (2 z)}}{ln ( 3 )}) : - \frac{3 ^{c o s (2 z)}}{2 ln ( 3 )}

= - \frac{3 ^{c o s (2 z)}}{2 ln ( 3 )}

Add a constant to the solution

= - \frac{3 ^{c o s (2 z)}}{2 ln ( 3 )} + C

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What is the integral of (3^{cos(2z)})/(csc(2z)) ?
The integral of (3^{cos(2z)})/(csc(2z)) is -(3^{cos(2z)})/(2ln(3))+C