What is the integral of (6+3z)cos(1+4z) ?

The integral of (6+3z)cos(1+4z) is 3/2 sin(1+4z)+3(1/4 zsin(1+4z)+1/16 cos(1+4z))+C

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integral of (6+3z)cos(1+4z)

\int (6 + 3 z) cos (1 + 4 z) d z

\frac{3}{2} sin (1 + 4 z) + 3 (\frac{1}{4} z sin (1 + 4 z) + \frac{1}{16} cos (1 + 4 z)) + C

Solution steps

\int (6 + 3 z) cos (1 + 4 z) d z

Expand

(6 + 3 z) cos (1 + 4 z) : 6 cos (1 + 4 z) + 3 z cos (1 + 4 z)

= \int 6 cos (1 + 4 z) + 3 z cos (1 + 4 z) d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 6 cos (1 + 4 z) d z + \int 3 z cos (1 + 4 z) d z

\int 6 cos (1 + 4 z) d z = \frac{3}{2} sin (1 + 4 z)

\int 3 z cos (1 + 4 z) d z = 3 (\frac{1}{4} z sin (1 + 4 z) + \frac{1}{16} cos (1 + 4 z))

= \frac{3}{2} sin (1 + 4 z) + 3 (\frac{1}{4} z sin (1 + 4 z) + \frac{1}{16} cos (1 + 4 z))

Add a constant to the solution

= \frac{3}{2} sin (1 + 4 z) + 3 (\frac{1}{4} z sin (1 + 4 z) + \frac{1}{16} cos (1 + 4 z)) + C

\frac{d}{d x} (3 x^{2} + \frac{10 x}{3} - 2)

(- 5.5) (0.1)

\int \frac{1}{( t - 1 ) ^{2}} d t

\frac{d}{d x} (\frac{x - 7}{x + 7})

\frac{d}{d x} (\frac{x}{4 + x})

What is the integral of (6+3z)cos(1+4z) ?
The integral of (6+3z)cos(1+4z) is 3/2 sin(1+4z)+3(1/4 zsin(1+4z)+1/16 cos(1+4z))+C