What is the integral of sin^6(θ) ?

The integral of sin^6(θ) is -(cos(θ)sin^5(θ))/6+5/6 (-1/4 sin^3(θ)cos(θ)+3/8 (θ-1/2 sin(2θ)))+C

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integral of sin^6(θ)

\int sin^{6} (θ) d θ

- \frac{cos ( θ ) sin ^{5} ( θ )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (θ) cos (θ) + \frac{3}{8} (θ - \frac{1}{2} sin (2 θ))) + C

Solution steps

\int sin^{6} (θ) d θ

Apply Integral Reduction

= - \frac{cos ( θ ) sin ^{5} ( θ )}{6} + \frac{5}{6} \cdot \int sin^{4} (θ) d θ

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

= - \frac{cos ( θ ) sin ^{5} ( θ )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (θ) cos (θ) + \frac{3}{8} (θ - \frac{1}{2} sin (2 θ)))

Add a constant to the solution

= - \frac{cos ( θ ) sin ^{5} ( θ )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (θ) cos (θ) + \frac{3}{8} (θ - \frac{1}{2} sin (2 θ))) + C

x \to \infty lim (x^{2} - 3 x)

\frac{d}{d x} (50 sin (2000 x))

x = y^{2} - 2, x = - y^{2} + 2

\int \frac{8 x}{x - 1} d x

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = - \frac{25}{3}

What is the integral of sin^6(θ) ?
The integral of sin^6(θ) is -(cos(θ)sin^5(θ))/6+5/6 (-1/4 sin^3(θ)cos(θ)+3/8 (θ-1/2 sin(2θ)))+C