What is the integral of sin^4(7θ) ?

The integral of sin^4(7θ) is 1/7 (-1/4 sin^3(7θ)cos(7θ)+3/8 (7θ-1/2 sin(14θ)))+C

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integral of sin^4(7θ)

\int sin^{4} (7 θ) d θ

\frac{1}{7} (- \frac{1}{4} sin^{3} (7 θ) cos (7 θ) + \frac{3}{8} (7 θ - \frac{1}{2} sin (14 θ))) + C

Solution steps

\int sin^{4} (7 θ) d θ

Apply u-substitution

: \int sin^{4} (u) \frac{1}{7} d u

= \int sin^{4} (u) \frac{1}{7} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{7} \cdot \int sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{7} (- \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{1}{7} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = 7 θ

= \frac{1}{7} (- \frac{cos ( 7 θ ) sin ^{3} ( 7 θ )}{4} + \frac{3}{4} \cdot \frac{1}{2} (7 θ - \frac{1}{2} sin (2 \cdot 7 θ)))

Simplify

\frac{1}{7} (- \frac{cos ( 7 θ ) sin ^{3} ( 7 θ )}{4} + \frac{3}{4} \cdot \frac{1}{2} (7 θ - \frac{1}{2} sin (2 \cdot 7 θ))) : \frac{1}{7} (- \frac{1}{4} sin^{3} (7 θ) cos (7 θ) + \frac{3}{8} (7 θ - \frac{1}{2} sin (14 θ)))

= \frac{1}{7} (- \frac{1}{4} sin^{3} (7 θ) cos (7 θ) + \frac{3}{8} (7 θ - \frac{1}{2} sin (14 θ)))

Add a constant to the solution

= \frac{1}{7} (- \frac{1}{4} sin^{3} (7 θ) cos (7 θ) + \frac{3}{8} (7 θ - \frac{1}{2} sin (14 θ))) + C

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What is the integral of sin^4(7θ) ?
The integral of sin^4(7θ) is 1/7 (-1/4 sin^3(7θ)cos(7θ)+3/8 (7θ-1/2 sin(14θ)))+C