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

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

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integral of-sin^4(x)

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integral Reduction

= - (- \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))

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x ^{2} - 3 x - 5}{x} d x

\frac{d}{d x} (x^{2} + \frac{432}{x})

\int \frac{1}{3 + x} d x

f (x) = ln (6 x^{3} + e^{x})

\frac{d}{d x} (\frac{1}{( x ^{2} - 3 x ) ^{2}})

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