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

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

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integral of sin^2(x)-sin^4(x)

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

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

x \to - 1 lim (\frac{x + 1}{x + 1})

\int 3 sin (x) - 2 cos (x) d x

y = 70 x^{- 4}

\int_{0}^{1} sin^{2} (x) d x

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