What is the integral of sin^4(5x) ?

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

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integral of sin^4(5x)

\int sin^{4} (5 x) d x

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

Solution steps

\int sin^{4} (5 x) d x

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 x

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

Simplify

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

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

Add a constant to the solution

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

2 y y^{^{'}} + 6 = y^{2} + 6 x, y (0) = 7

x \to 1 + lim (\frac{- 1}{x ^{2} - 1})

\int \frac{x}{( x ^{2} + 4 ) ^{2}} d x

- \frac{11}{s ^{4}}

n = 1 \sum \infty \frac{n x ^{n}}{5 ^{n}}

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