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

1 - cos^{2} (x)

\int x^{2} (x - 6) d x

\int_{0}^{2 π} sin^{2} (15 x) d x

x \to 4 lim (\frac{x ^{2} + 16 - 64}{x})

n = 0 \sum \infty ((\frac{1}{3})^{n})^{2}

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