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

The integral of (x^2-2x+1)cos(x) is x^2sin(x)+2xcos(x)-2xsin(x)-sin(x)-2cos(x)+C

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integral of (x^2-2x+1)cos(x)

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

x^{2} sin (x) + 2 x cos (x) - 2 x sin (x) - sin (x) - 2 cos (x) + C

Solution steps

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

Expand

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

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

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

\int 2 x cos (x) d x = 2 (x sin (x) + cos (x))

\int cos (x) d x = sin (x)

= x^{2} sin (x) - 2 (- x cos (x) + sin (x)) - 2 (x sin (x) + cos (x)) + sin (x)

Simplify

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

= x^{2} sin (x) + 2 x cos (x) - 2 x sin (x) - sin (x) - 2 cos (x)

Add a constant to the solution

= x^{2} sin (x) + 2 x cos (x) - 2 x sin (x) - sin (x) - 2 cos (x) + C

\frac{x}{x ^{2} + 1}

\int 4 e^{- x} cos (x) d x

\int \frac{3 cos ( x )}{- 2 + 10 sin ( x )} d x

\int e^{- t} \cdot sin (t) d t

\int \frac{ln ^{2} ( x ^{4} )}{x} d x

What is the integral of (x^2-2x+1)cos(x) ?
The integral of (x^2-2x+1)cos(x) is x^2sin(x)+2xcos(x)-2xsin(x)-sin(x)-2cos(x)+C