What is the integral of sin(3z)cos(2)(3z) ?

The integral of sin(3z)cos(2)(3z) is -cos(2)zcos(3z)+1/3 cos(2)sin(3z)+C

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integral of sin(3z)cos(2)(3z)

\int sin (3 z) cos (2) (3 z) d z

- cos (2) z cos (3 z) + \frac{1}{3} cos (2) sin (3 z) + C

Solution steps

\int sin (3 z) cos (2) \cdot 3 z d z

Take the constant out:

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

= cos (2) \cdot 3 \cdot \int sin (3 z) z d z

Apply Integration By Parts

: - \frac{1}{3} z cos (3 z) - \int - \frac{1}{3} cos (3 z) d z

= cos (2) \cdot 3 (- \frac{1}{3} z cos (3 z) - \int - \frac{1}{3} cos (3 z) d z)

\int - \frac{1}{3} cos (3 z) d z = - \frac{1}{9} sin (3 z)

= cos (2) \cdot 3 (- \frac{1}{3} z cos (3 z) - (- \frac{1}{9} sin (3 z)))

Simplify

cos (2) \cdot 3 (- \frac{1}{3} z cos (3 z) - (- \frac{1}{9} sin (3 z))) : - cos (2) z cos (3 z) + \frac{1}{3} cos (2) sin (3 z)

= - cos (2) z cos (3 z) + \frac{1}{3} cos (2) sin (3 z)

Add a constant to the solution

= - cos (2) z cos (3 z) + \frac{1}{3} cos (2) sin (3 z) + C

f (t) = t cos (2 t) e^{- t}, at t = 3

(1 + x^{2} + y^{2} + x^{2} y^{2}) d y = y^{2} d x

\int \frac{1}{3 x - 2} d x

f (x) = (9 x^{3} + 3 x^{2})^{4}

\frac{d}{d x} (6 + tan^{2} (x))

What is the integral of sin(3z)cos(2)(3z) ?
The integral of sin(3z)cos(2)(3z) is -cos(2)zcos(3z)+1/3 cos(2)sin(3z)+C