What is the integral of (x-5)sin(x^2-10x+1) ?

The integral of (x-5)sin(x^2-10x+1) is -1/2 cos(x^2-10x+1)+C

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integral of (x-5)sin(x^2-10x+1)

\int (x - 5) sin (x^{2} - 10 x + 1) d x

- \frac{1}{2} cos (x^{2} - 10 x + 1) + C

Solution steps

\int (x - 5) sin (x^{2} - 10 x + 1) d x

Apply u-substitution

: \int \frac{sin ( u )}{2} d u

= \int \frac{sin ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{2} (- cos (u))

Substitute back

u = x^{2} - 10 x + 1

= \frac{1}{2} (- cos (x^{2} - 10 x + 1))

Simplify

= - \frac{1}{2} cos (x^{2} - 10 x + 1)

Add a constant to the solution

= - \frac{1}{2} cos (x^{2} - 10 x + 1) + C

sin (4 t) + cos (4 t)

\frac{1}{s ( s + 1 ) ^{4}}

\int 3 tan (5 x) sec^{2} (5 x) d x

h \to 0 lim (5 x^{2} + 8 y^{2})

n = 0 \sum \infty (2 n)! (\frac{x}{3})^{n}

What is the integral of (x-5)sin(x^2-10x+1) ?
The integral of (x-5)sin(x^2-10x+1) is -1/2 cos(x^2-10x+1)+C