What is the integral of (sin(θ))^4 ?

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

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integral of (sin(θ))^4

\int (sin (θ))^{4} d θ

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

Solution steps

\int (sin (θ))^{4} d θ

Simplify

= \int sin^{4} (θ) d θ

Apply Integral Reduction

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

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

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

Simplify

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

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

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