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

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

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integral of 5(sin(x))^4

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

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

Solution steps

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

Take the constant out:

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

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

Simplify

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

Apply Integral Reduction

= 5 (- \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))

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

Simplify

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

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

Add a constant to the solution

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

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