What is the integral of 3sin(y)+1 ?

The integral of 3sin(y)+1 is -3cos(y)+y+C

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integral of 3sin(y)+1

\int 3 sin (y) + 1 d y

- 3 cos (y) + y + C

Solution steps

\int 3 sin (y) + 1 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 sin (y) d y + \int 1 d y

\int 3 sin (y) d y = - 3 cos (y)

\int 1 d y = y

= - 3 cos (y) + y

Add a constant to the solution

= - 3 cos (y) + y + C

\frac{d}{d x} ((x^{2} - 1)^{2})

(e^{x} ln (y)) d x + (\frac{e ^{2 x}}{2 y}) d y = 0, y (0) = e

h (x) = \frac{6}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}}

y^{^{''}} - 4 y = 0, y (2) = 2, y^{^{'}} (- 1) = 1

x y^{^{'}} - 4 y = 3 x