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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= - \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 - (1 - t)^{2}))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (cos (x π))

\int e^{1.25 x} - 4 x d x

\frac{\partial}{\partial x} (6 x^{\frac{1}{3}} y^{\frac{1}{2}})

\int \frac{e ^{x}}{1 + 25 e ^{2 x}} d x

f (x) = 2 x + \frac{3}{x}, at x = - 1

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