What is the integral of sin^4(5θ) ?

The integral of sin^4(5θ) is 1/5 (-1/4 sin^3(5θ)cos(5θ)+3/8 (5θ-1/2 sin(10θ)))+C

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integral of sin^4(5θ)

\int sin^{4} (5 θ) d θ

\frac{1}{5} (- \frac{1}{4} sin^{3} (5 θ) cos (5 θ) + \frac{3}{8} (5 θ - \frac{1}{2} sin (10 θ))) + C

Solution steps

\int sin^{4} (5 θ) d θ

Apply u-substitution

: \int sin^{4} (u) \frac{1}{5} d u

= \int sin^{4} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{5} (- \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}{5} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = 5 θ

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

Simplify

\frac{1}{5} (- \frac{cos ( 5 θ ) sin ^{3} ( 5 θ )}{4} + \frac{3}{4} \cdot \frac{1}{2} (5 θ - \frac{1}{2} sin (2 \cdot 5 θ))) : \frac{1}{5} (- \frac{1}{4} sin^{3} (5 θ) cos (5 θ) + \frac{3}{8} (5 θ - \frac{1}{2} sin (10 θ)))

= \frac{1}{5} (- \frac{1}{4} sin^{3} (5 θ) cos (5 θ) + \frac{3}{8} (5 θ - \frac{1}{2} sin (10 θ)))

Add a constant to the solution

= \frac{1}{5} (- \frac{1}{4} sin^{3} (5 θ) cos (5 θ) + \frac{3}{8} (5 θ - \frac{1}{2} sin (10 θ))) + C

y = (cos (3 x))^{x}

x \to 0 lim (\frac{sin ( 4 x )}{x})

\frac{d}{d x} (A cos (2 x))

\int sin (nπ x) d x

\frac{d}{d x} ((x + 1) (x + 3))

What is the integral of sin^4(5θ) ?
The integral of sin^4(5θ) is 1/5 (-1/4 sin^3(5θ)cos(5θ)+3/8 (5θ-1/2 sin(10θ)))+C