What is the integral of 32sin^4(2x) ?

The integral of 32sin^4(2x) is 16(-1/4 sin^3(2x)cos(2x)+3/8 (2x-1/2 sin(4x)))+C

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integral of 32sin^4(2x)

\int 32 sin^{4} (2 x) d x

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

Solution steps

\int 32 sin^{4} (2 x) d x

Take the constant out:

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

= 32 \cdot \int sin^{4} (2 x) d x

Apply u-substitution

: \int sin^{4} (u) \frac{1}{2} d u

= 32 \cdot \int sin^{4} (u) \frac{1}{2} d u

Take the constant out:

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

= 32 \cdot \frac{1}{2} \cdot \int sin^{4} (u) d u

Apply Integral Reduction

= 32 \cdot \frac{1}{2} (- \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))

= 32 \cdot \frac{1}{2} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = 2 x

= 32 \cdot \frac{1}{2} (- \frac{cos ( 2 x ) sin ^{3} ( 2 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (2 x - \frac{1}{2} sin (2 \cdot 2 x)))

Simplify

32 \cdot \frac{1}{2} (- \frac{cos ( 2 x ) sin ^{3} ( 2 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (2 x - \frac{1}{2} sin (2 \cdot 2 x))) : 16 (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (4 x)))

= 16 (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (4 x)))

Add a constant to the solution

= 16 (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (4 x))) + C

\frac{d}{d y} ((x^{2} + y^{2}) arctan (\frac{y}{x}))

e^{x}, x = 2

\int \frac{x}{7} d x

y^{^{'}} - 5 y = 10 x

\frac{1}{( s - 4 ) ^{2}}

What is the integral of 32sin^4(2x) ?
The integral of 32sin^4(2x) is 16(-1/4 sin^3(2x)cos(2x)+3/8 (2x-1/2 sin(4x)))+C