What is the integral of (sin(6x)+cos(1/3)x) ?

The integral of (sin(6x)+cos(1/3)x) is -1/6 cos(6x)+1/2 cos(1/3)x^2+C

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integral of (sin(6x)+cos(1/3)x)

\int (sin (6 x) + cos (\frac{1}{3}) x) d x

- \frac{1}{6} cos (6 x) + \frac{1}{2} cos (\frac{1}{3}) x^{2} + C

Solution steps

\int sin (6 x) + cos (\frac{1}{3}) 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 sin (6 x) d x + \int cos (\frac{1}{3}) x d x

\int sin (6 x) d x = - \frac{1}{6} cos (6 x)

\int cos (\frac{1}{3}) x d x = \frac{1}{2} cos (\frac{1}{3}) x^{2}

= - \frac{1}{6} cos (6 x) + \frac{1}{2} cos (\frac{1}{3}) x^{2}

Add a constant to the solution

= - \frac{1}{6} cos (6 x) + \frac{1}{2} cos (\frac{1}{3}) x^{2} + C

\int e^{- \frac{2}{3} x} d x

\int 4 x^{3} - x d x

\int \frac{8 x ^{2} + x}{( 16 x ^{3} + 3 x ^{2} ) ^{2}} d x

\int (4 x^{2} - 3)^{2} d x

What is the integral of (sin(6x)+cos(1/3)x) ?
The integral of (sin(6x)+cos(1/3)x) is -1/6 cos(6x)+1/2 cos(1/3)x^2+C