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

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

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integral of 3cos^3(x)

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

3 sin (x) - sin^{3} (x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

= 3 \cdot \int (1 - sin^{2} (x)) cos (x) d x

Apply u-substitution

: \int 1 - u^{2} d u

= 3 \cdot \int 1 - u^{2} d u

Apply the Sum Rule:

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

= 3 (\int 1 d u - \int u^{2} d u)

\int 1 d u = u

\int u^{2} d u = \frac{u ^{3}}{3}

= 3 (u - \frac{u ^{3}}{3})

Substitute back

u = sin (x)

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

Simplify

3 (sin (x) - \frac{sin ^{3} ( x )}{3}) : 3 sin (x) - sin^{3} (x)

= 3 sin (x) - sin^{3} (x)

Add a constant to the solution

= 3 sin (x) - sin^{3} (x) + C

\frac{d}{d x} (\frac{4 x ^{2} + 8 x + 6}{x})

\int_{0}^{1} x 4 + 9 x^{2} d x

\int e^{2 x} \cdot sin (e^{x}) d x

\frac{\partial}{\partial x} (\frac{5 x yz}{x ^{2} + y ^{2} + z ^{2}})

y^{^{'}} + 2 y = 10 sin (t)

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