What is the integral of 6sin^4(x) ?

The integral of 6sin^4(x) is 6(-1/4 sin^3(x)cos(x)+3/8 (x-1/2 sin(2x)))+C

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integral of 6sin^4(x)

\int 6 sin^{4} (x) d x

6 (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))) + C

Solution steps

\int 6 sin^{4} (x) d x

Take the constant out:

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

= 6 \cdot \int sin^{4} (x) d x

Apply Integral Reduction

= 6 (- \frac{cos ( x ) sin ^{3} ( x )}{4} + \frac{3}{4} \cdot \int sin^{2} (x) d x)

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

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

Simplify

6 (- \frac{cos ( x ) sin ^{3} ( x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x))) : 6 (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))

= 6 (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))

Add a constant to the solution

= 6 (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))) + C

\frac{d}{d x} ((12 x^{2} - 12) (9 - 4 x))

\frac{\partial}{\partial x} (x^{2} - 2 x y + 3 y^{2} x^{3})

y = ln (\frac{x ^{2} - 1}{x ^{2} + 1})

\frac{d y}{d x} + 12 y = 6

- 10 e^{- 5 x^{2}} x

What is the integral of 6sin^4(x) ?
The integral of 6sin^4(x) is 6(-1/4 sin^3(x)cos(x)+3/8 (x-1/2 sin(2x)))+C