What is the integral of (sqrt(2x)-cos(6x)) ?

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

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integral of (sqrt(2x)-cos(6x))

\int (2 x - cos (6 x)) d x

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

Solution steps

\int 2 x - cos (6 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 2 x d x - \int cos (6 x) d x

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

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

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

Add a constant to the solution

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

x \to \infty lim (0. 8^{x})

\int 3 x^{2} 2 x^{3} - 5 d x

f (x) = 4 - x^{2}, at x = 1.154

x^{3} - 3 x^{2} + 2 x - 2, at x = 0

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 2 y})

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