What is the integral of (x-1)^2(sin(x))^2 ?

The integral of (x-1)^2(sin(x))^2 is

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integral of (x-1)^2(sin(x))^2

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

\frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - x^{2} + \frac{x ^{2}}{2} - \frac{1}{4} x cos (2 x) + \frac{1}{2} x sin (2 x) + \frac{1}{2} x - \frac{1}{8} sin (2 x) + \frac{1}{4} cos (2 x) + C

Solution steps

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

Simplify

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

Expand

sin^{2} (x) (x - 1)^{2} : x^{2} sin^{2} (x) - 2 x sin^{2} (x) + sin^{2} (x)

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

\int x^{2} sin^{2} (x) d x = \frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - \frac{1}{4} x cos (2 x) + \frac{1}{8} sin (2 x)

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

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

= \frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - \frac{1}{4} x cos (2 x) + \frac{1}{8} sin (2 x) - (x^{2} - \frac{1}{2} x sin (2 x) - \frac{x ^{2}}{2} - \frac{1}{4} cos (2 x)) + \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Simplify

\frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - \frac{1}{4} x cos (2 x) + \frac{1}{8} sin (2 x) - (x^{2} - \frac{1}{2} x sin (2 x) - \frac{x ^{2}}{2} - \frac{1}{4} cos (2 x)) + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : \frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - x^{2} + \frac{x ^{2}}{2} - \frac{1}{4} x cos (2 x) + \frac{1}{2} x sin (2 x) + \frac{1}{2} x - \frac{1}{8} sin (2 x) + \frac{1}{4} cos (2 x)

= \frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - x^{2} + \frac{x ^{2}}{2} - \frac{1}{4} x cos (2 x) + \frac{1}{2} x sin (2 x) + \frac{1}{2} x - \frac{1}{8} sin (2 x) + \frac{1}{4} cos (2 x)

Add a constant to the solution

= \frac{1}{3} x^{3} sin^{2} (x) + \frac{1}{6} x^{3} cos (2 x) - \frac{1}{4} x^{2} sin (2 x) - x^{2} + \frac{x ^{2}}{2} - \frac{1}{4} x cos (2 x) + \frac{1}{2} x sin (2 x) + \frac{1}{2} x - \frac{1}{8} sin (2 x) + \frac{1}{4} cos (2 x) + C

f (t) = 1.3 e^{0.7 t}

\int x \cdot e^{1 - x} d x

(2 x - 5)^{4} (8 x^{2} - 5)^{- 3}

\frac{\partial}{\partial y} (y e^{- x y})

\frac{d}{d x} (ln (\frac{e ^{x}}{1 + e ^{x}}))

What is the integral of (x-1)^2(sin(x))^2 ?
The integral of (x-1)^2(sin(x))^2 is