What is the integral of (2x^2+3x-5)sin(x) ?

The integral of (2x^2+3x-5)sin(x) is -2x^2cos(x)+4xsin(x)-3xcos(x)+9cos(x)+3sin(x)+C

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integral of (2x^2+3x-5)sin(x)

\int (2 x^{2} + 3 x - 5) sin (x) d x

- 2 x^{2} cos (x) + 4 x sin (x) - 3 x cos (x) + 9 cos (x) + 3 sin (x) + C

Solution steps

\int (2 x^{2} + 3 x - 5) sin (x) d x

Expand

(2 x^{2} + 3 x - 5) sin (x) : 2 x^{2} sin (x) + 3 x sin (x) - 5 sin (x)

= \int 2 x^{2} sin (x) + 3 x sin (x) - 5 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 2 x^{2} sin (x) d x + \int 3 x sin (x) d x - \int 5 sin (x) d x

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

\int 3 x sin (x) d x = 3 (- x cos (x) + sin (x))

\int 5 sin (x) d x = - 5 cos (x)

= 2 (- x^{2} cos (x) + 2 (x sin (x) + cos (x))) + 3 (- x cos (x) + sin (x)) - (- 5 cos (x))

Simplify

2 (- x^{2} cos (x) + 2 (x sin (x) + cos (x))) + 3 (- x cos (x) + sin (x)) - (- 5 cos (x)) : - 2 x^{2} cos (x) + 4 x sin (x) - 3 x cos (x) + 9 cos (x) + 3 sin (x)

= - 2 x^{2} cos (x) + 4 x sin (x) - 3 x cos (x) + 9 cos (x) + 3 sin (x)

Add a constant to the solution

= - 2 x^{2} cos (x) + 4 x sin (x) - 3 x cos (x) + 9 cos (x) + 3 sin (x) + C

f (x) = 3 x^{3} - x^{2} + 7, at x = 2

6 x - x^{2}

\int_{1}^{5} \frac{x}{x - 1} d x

\int \frac{16 x ^{2}}{( x ^{2} + 9 ) ^{5}} d x

\int - 2 csc^{2} (x) cot (x) d x

What is the integral of (2x^2+3x-5)sin(x) ?
The integral of (2x^2+3x-5)sin(x) is -2x^2cos(x)+4xsin(x)-3xcos(x)+9cos(x)+3sin(x)+C