What is the integral of 6x^2arcos(x) ?

The integral of 6x^2arcos(x) is 6ar(x^2sin(x)-2(-xcos(x)+sin(x)))+C

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integral of 6x^2arcos(x)

\int 6 x^{2} a r cos (x) d x

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

Solution steps

\int 6 x^{2} a r cos (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 a r \cdot \int x^{2} cos (x) d x

Apply Integration By Parts

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

= 6 a r (x^{2} sin (x) - \int 2 x sin (x) d x)

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

= 6 a r (x^{2} sin (x) - 2 (- x cos (x) + sin (x)))

Add a constant to the solution

= 6 a r (x^{2} sin (x) - 2 (- x cos (x) + sin (x))) + C

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What is the integral of 6x^2arcos(x) ?
The integral of 6x^2arcos(x) is 6ar(x^2sin(x)-2(-xcos(x)+sin(x)))+C