What is the integral of cos(x)+1/9 x ?

The integral of cos(x)+1/9 x is sin(x)+(x^2)/(18)+C

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integral of cos(x)+1/9 x

\int cos (x) + \frac{1}{9} x d x

sin (x) + \frac{x ^{2}}{18} + C

Solution steps

\int cos (x) + \frac{1}{9} 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 cos (x) d x + \int \frac{1}{9} x d x

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

\int \frac{1}{9} x d x = \frac{x ^{2}}{18}

= sin (x) + \frac{x ^{2}}{18}

Add a constant to the solution

= sin (x) + \frac{x ^{2}}{18} + C

\int_{0}^{5} \frac{x}{9 - x} d x

y = 4 - x^{2}, y = - x + 2

\int \frac{x - 1}{2 x + 4} d x

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

\int cos^{4} (4 x) d x

What is the integral of cos(x)+1/9 x ?
The integral of cos(x)+1/9 x is sin(x)+(x^2)/(18)+C