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

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

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integral of sin^4(t)

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

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

Solution steps

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Add a constant to the solution

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

x (\frac{d y}{d x}) - 3 y = x^{3}, y (1) = 1

\frac{d y}{d x} = \frac{y}{x} - \frac{x ^{2}}{y ^{2}}

x \to 0 + lim (\frac{ln ( x )}{x ^{2}})

\frac{d y}{d t} = 5 y + y^{5}

x - \frac{1}{3 x ^{3}}

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