What is the integral of 5^{cos(3x)}sin(3x) ?

The integral of 5^{cos(3x)}sin(3x) is -(5^{cos(3x)})/(3ln(5))+C

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integral of 5^{cos(3x)}sin(3x)

\int 5^{c o s (3 x)} sin (3 x) d x

- \frac{5 ^{c o s (3 x)}}{3 ln ( 5 )} + C

Solution steps

\int 5^{c o s (3 x)} sin (3 x) d x

Apply u-substitution

: \int 5^{c o s (u)} sin (u) \frac{1}{3} d u

= \int 5^{c o s (u)} sin (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int 5^{c o s (u)} sin (u) d u

Apply u-substitution

: \int - 5^{v} d v

= \frac{1}{3} \cdot \int - 5^{v} d v

Take the constant out:

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

= \frac{1}{3} (- \int 5^{v} d v)

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

= \frac{1}{3} (- \frac{5 ^{v}}{ln ( 5 )})

Substitute back

= \frac{1}{3} (- \frac{5 ^{c o s (3 x)}}{ln ( 5 )})

Simplify

\frac{1}{3} (- \frac{5 ^{c o s (3 x)}}{ln ( 5 )}) : - \frac{5 ^{c o s (3 x)}}{3 ln ( 5 )}

= - \frac{5 ^{c o s (3 x)}}{3 ln ( 5 )}

Add a constant to the solution

= - \frac{5 ^{c o s (3 x)}}{3 ln ( 5 )} + C

f (x) = 9 e^{x}

e^{2 x} - x

10 y^{^{''}} - 10 y^{^{'}} + 125 y = 0

x \to \infty lim ((1 + \frac{1}{6 x})^{5 x})

3 t^{2}

What is the integral of 5^{cos(3x)}sin(3x) ?
The integral of 5^{cos(3x)}sin(3x) is -(5^{cos(3x)})/(3ln(5))+C