What is the integral of 8sin^4(x) ?

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

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integral of 8sin^4(x)

\int 8 sin^{4} (x) d x

- 2 sin^{3} (x) cos (x) + 3 x - \frac{3}{2} sin (2 x) + C

Solution steps

\int 8 sin^{4} (x) d x

Take the constant out:

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

= 8 \cdot \int sin^{4} (x) d x

Apply Integral Reduction

= 8 (- \frac{cos ( x ) sin ^{3} ( x )}{4} + \frac{3}{4} \cdot \int sin^{2} (x) d x)

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

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

Simplify

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

= - 2 sin^{3} (x) cos (x) + 3 x - \frac{3}{2} sin (2 x)

Add a constant to the solution

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

\int \frac{x}{x ^{2}} d x

x \to 0 lim (\frac{x}{cot ( 5 x )})

\int \frac{1}{1 + v ^{2}} d v

x \to 2 lim (- 4)

\int x e^{- x (1 + y)} d x

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