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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

\int x z e^{yz} d y

\int (2 x - 4)^{13} d x

\int - 2 e^{- 2 t} d t

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