What is the integral of 3sin(3x)+2cos(4x) ?

The integral of 3sin(3x)+2cos(4x) is -cos(3x)+1/2 sin(4x)+C

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integral of 3sin(3x)+2cos(4x)

\int 3 sin (3 x) + 2 cos (4 x) d x

- cos (3 x) + \frac{1}{2} sin (4 x) + C

Solution steps

\int 3 sin (3 x) + 2 cos (4 x) d x

Apply the Sum Rule:

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

= \int 3 sin (3 x) d x + \int 2 cos (4 x) d x

\int 3 sin (3 x) d x = - cos (3 x)

\int 2 cos (4 x) d x = \frac{1}{2} sin (4 x)

= - cos (3 x) + \frac{1}{2} sin (4 x)

Add a constant to the solution

= - cos (3 x) + \frac{1}{2} sin (4 x) + C

y^{^{'''}} + 2 y^{^{''}} - y^{^{'}} - 2 y = sin (3 t)

\frac{3 x}{x}

\int - x^{3} - 8 x + 5 d x

x \to 0 lim (\frac{1}{x + x ^{5}})

2 x^{2} - x^{3}, y = 0

What is the integral of 3sin(3x)+2cos(4x) ?
The integral of 3sin(3x)+2cos(4x) is -cos(3x)+1/2 sin(4x)+C