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

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

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integral of-sin(x)+3cos(x)-1x

\int - sin (x) + 3 cos (x) - 1 x d x

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

Solution steps

\int - sin (x) + 3 cos (x) - 1 \cdot 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 (x) d x + \int 3 cos (x) d x - \int 1 \cdot x d x

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

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

\int 1 \cdot x d x = \frac{x ^{2}}{2}

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

Simplify

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

Add a constant to the solution

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

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