What is the integral of 16sin^4(x) ?

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

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integral of 16sin^4(x)

\int 16 sin^{4} (x) d x

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

Solution steps

\int 16 sin^{4} (x) d x

Take the constant out:

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

= 16 \cdot \int sin^{4} (x) d x

Apply Integral Reduction

= 16 (- \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))

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

Simplify

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

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

Add a constant to the solution

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

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