What is the integral of sin^4(u) ?

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

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integral of sin^4(u)

\int sin^{4} (u) d u

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

Solution steps

\int sin^{4} (u) d u

Apply Integral Reduction

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (8 x - 2 x^{3})

\frac{d}{d x} (\frac{3 x}{1 - x})

\int e^{\frac{1}{x}} d x

d er i v a t i v e (5 + 6 x)^{2}

\frac{\partial}{\partial x} (ln (x y^{2} z^{3}))

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