What is the integral of 4-4sin(x)+sin^2(x) ?

The integral of 4-4sin(x)+sin^2(x) is 9/2 x-1/4 sin(2x)+4cos(x)+C

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integral of 4-4sin(x)+sin^2(x)

\int 4 - 4 sin (x) + sin^{2} (x) d x

\frac{9}{2} x - \frac{1}{4} sin (2 x) + 4 cos (x) + C

Solution steps

\int 4 - 4 sin (x) + sin^{2} (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 4 d x - \int 4 sin (x) d x + \int sin^{2} (x) d x

\int 4 d x = 4 x

\int 4 sin (x) d x = - 4 cos (x)

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

= 4 x - (- 4 cos (x)) + \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Simplify

4 x - (- 4 cos (x)) + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : \frac{9}{2} x - \frac{1}{4} sin (2 x) + 4 cos (x)

= \frac{9}{2} x - \frac{1}{4} sin (2 x) + 4 cos (x)

Add a constant to the solution

= \frac{9}{2} x - \frac{1}{4} sin (2 x) + 4 cos (x) + C

\int 64 - 3 x^{2} d x

\frac{\partial}{\partial x} (x^{2} e^{2})

x \to 1 lim (\frac{a x}{3})

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

x \to 2 lim (\frac{x ^{3} - 4}{x - 2})

What is the integral of 4-4sin(x)+sin^2(x) ?
The integral of 4-4sin(x)+sin^2(x) is 9/2 x-1/4 sin(2x)+4cos(x)+C