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

The integral of x^2cos(2x-1) is -1/4 (2x^2sin(-2x+1)-sin(-2x+1)-2xcos(-2x+1))+C

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

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

- \frac{1}{4} (2 x^{2} sin (- 2 x + 1) - sin (- 2 x + 1) - 2 x cos (- 2 x + 1)) + C

Solution steps

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

Factor

2 x - 1 : - (- 2 x + 1)

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

Simplify

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

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

Apply u-substitution

: \int - \frac{cos ( u ) ( u - 1 ) ^{2}}{8} d u

= \int - \frac{cos ( u ) ( u - 1 ) ^{2}}{8} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int cos (u) (u - 1)^{2} d u

Apply Integration By Parts

: (u - 1)^{2} sin (u) - \int 2 (u - 1) sin (u) d u

= - \frac{1}{8} ((u - 1)^{2} sin (u) - \int 2 (u - 1) sin (u) d u)

\int 2 (u - 1) sin (u) d u = 2 (- cos (u) (u - 1) + sin (u))

= - \frac{1}{8} ((u - 1)^{2} sin (u) - 2 (- cos (u) (u - 1) + sin (u)))

Substitute back

u = 1 - 2 x

= - \frac{1}{8} ((1 - 2 x - 1)^{2} sin (1 - 2 x) - 2 (- cos (1 - 2 x) (1 - 2 x - 1) + sin (1 - 2 x)))

Simplify

- \frac{1}{8} ((1 - 2 x - 1)^{2} sin (1 - 2 x) - 2 (- cos (1 - 2 x) (1 - 2 x - 1) + sin (1 - 2 x))) : - \frac{1}{4} (2 x^{2} sin (- 2 x + 1) - sin (- 2 x + 1) - 2 x cos (- 2 x + 1))

= - \frac{1}{4} (2 x^{2} sin (- 2 x + 1) - sin (- 2 x + 1) - 2 x cos (- 2 x + 1))

Add a constant to the solution

= - \frac{1}{4} (2 x^{2} sin (- 2 x + 1) - sin (- 2 x + 1) - 2 x cos (- 2 x + 1)) + C

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What is the integral of x^2cos(2x-1) ?
The integral of x^2cos(2x-1) is -1/4 (2x^2sin(-2x+1)-sin(-2x+1)-2xcos(-2x+1))+C