What is the integral of (sin(x)+7cos(x)-1) ?

The integral of (sin(x)+7cos(x)-1) is -cos(x)+7sin(x)-x+C

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integral of (sin(x)+7cos(x)-1)

\int (sin (x) + 7 cos (x) - 1) d x

- cos (x) + 7 sin (x) - x + C

Solution steps

\int sin (x) + 7 cos (x) - 1 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 7 cos (x) d x - \int 1 d x

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

\int 7 cos (x) d x = 7 sin (x)

\int 1 d x = x

= - cos (x) + 7 sin (x) - x

Add a constant to the solution

= - cos (x) + 7 sin (x) - x + C

(t e)^{3} e^{- 2 (t + 1)}

n = 5 \sum \infty e^{5 - 2 n}

\frac{3}{s ^{2} - 2 s - 15}

\int (\frac{7}{x ^{2} + 1}) d x

\frac{d}{d x} (csc^{3} (x^{2}))

What is the integral of (sin(x)+7cos(x)-1) ?
The integral of (sin(x)+7cos(x)-1) is -cos(x)+7sin(x)-x+C