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

The integral of (3-cos^3(x))sin(x) is -3cos(x)+1/4 cos^4(x)+C

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integral of (3-cos^3(x))sin(x)

\int (3 - cos^{3} (x)) sin (x) d x

- 3 cos (x) + \frac{1}{4} cos^{4} (x) + C

Solution steps

\int (3 - cos^{3} (x)) sin (x) d x

Expand

(3 - cos^{3} (x)) sin (x) : 3 sin (x) - cos^{3} (x) sin (x)

= \int 3 sin (x) - cos^{3} (x) sin (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 3 sin (x) d x - \int cos^{3} (x) sin (x) d x

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

\int cos^{3} (x) sin (x) d x = - \frac{cos ^{4} ( x )}{4}

= - 3 cos (x) - (- \frac{cos ^{4} ( x )}{4})

Simplify

- 3 cos (x) - (- \frac{cos ^{4} ( x )}{4}) : - 3 cos (x) + \frac{1}{4} cos^{4} (x)

= - 3 cos (x) + \frac{1}{4} cos^{4} (x)

Add a constant to the solution

= - 3 cos (x) + \frac{1}{4} cos^{4} (x) + C

3 sin (4 t)

\frac{d}{d x} (cos (x^{l n (x^{2} + 1)}))

x \to 0 lim (\frac{x - x}{x})

8 x^{^{''}} + 8 x^{^{'}} + \frac{1}{4} x = 0, x^{^{'}} (0) = 1, x (0) = 2

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

What is the integral of (3-cos^3(x))sin(x) ?
The integral of (3-cos^3(x))sin(x) is -3cos(x)+1/4 cos^4(x)+C