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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Simplify

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

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

Apply u-substitution

: \int - \frac{cos ( v )}{2} d v

= \int - \frac{cos ( v )}{2} d v

Take the constant out:

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

= - \frac{1}{2} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= - \frac{1}{2} sin (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int (x - 2) sin (x) d x

\int \frac{x ^{4} + 1}{x ^{3}} d x

\int (2 x + 10)^{20} d x

\int sin (arctan (x)) d x

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