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

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

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integral of (sin^2(x))^2

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

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

Solution steps

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

Rewrite using trig identities

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

Apply Integral Reduction

= - \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))

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

Simplify

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

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

Add a constant to the solution

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

\frac{d ^{2} y}{d x ^{2}} + 4 y = 0

(f (x) (2 x))^{'}

x \to - 8 lim (x + 8)

\frac{\partial}{\partial x} (\frac{y}{z} + \frac{z}{x} - \frac{x z}{y})

\int \frac{x}{x ^{2} + 2 x + 5} d x

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