What is the integral of sin(x)(3x^2+x-1) ?

The integral of sin(x)(3x^2+x-1) is -3x^2cos(x)+6xsin(x)-xcos(x)+7cos(x)+sin(x)+C

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integral of sin(x)(3x^2+x-1)

\int sin (x) (3 x^{2} + x - 1) d x

- 3 x^{2} cos (x) + 6 x sin (x) - x cos (x) + 7 cos (x) + sin (x) + C

Solution steps

\int sin (x) (3 x^{2} + x - 1) d x

Expand

sin (x) (3 x^{2} + x - 1) : 3 x^{2} sin (x) + x sin (x) - sin (x)

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

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

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

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

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

Simplify

3 (- x^{2} cos (x) + 2 (x sin (x) + cos (x))) - x cos (x) + sin (x) - (- cos (x)) : - 3 x^{2} cos (x) + 6 x sin (x) - x cos (x) + 7 cos (x) + sin (x)

= - 3 x^{2} cos (x) + 6 x sin (x) - x cos (x) + 7 cos (x) + sin (x)

Add a constant to the solution

= - 3 x^{2} cos (x) + 6 x sin (x) - x cos (x) + 7 cos (x) + sin (x) + C

\int 3 x^{2} 5^{4 x} d x

\int \frac{143}{( x + 7 ) ^{2}} d x

\int \frac{3}{4 x ^{2} + 4} d x

\int 2 q d q

What is the integral of sin(x)(3x^2+x-1) ?
The integral of sin(x)(3x^2+x-1) is -3x^2cos(x)+6xsin(x)-xcos(x)+7cos(x)+sin(x)+C