What is the integral of-8xsin(4x^2+1) ?

The integral of-8xsin(4x^2+1) is cos(4x^2+1)+C

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integral of-8xsin(4x^2+1)

\int - 8 x sin (4 x^{2} + 1) d x

cos (4 x^{2} + 1) + C

Solution steps

\int - 8 x sin (4 x^{2} + 1) d x

Take the constant out:

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

= - 8 \cdot \int x sin (4 x^{2} + 1) d x

Apply u-substitution

: \int \frac{sin ( u )}{8} d u

= - 8 \cdot \int \frac{sin ( u )}{8} d u

Take the constant out:

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

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

Use the common integral:

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

= - 8 \cdot \frac{1}{8} (- cos (u))

Substitute back

u = 4 x^{2} + 1

= - 8 \cdot \frac{1}{8} (- cos (4 x^{2} + 1))

Simplify

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

= cos (4 x^{2} + 1)

Add a constant to the solution

= cos (4 x^{2} + 1) + C

x^{2} + 5 y \frac{d y}{d x} = 0

x \to - 1 lim ((9 x + 1)^{2})

\int 2 x y e^{x^{2}} d x

\int \frac{1}{x ( ln ( x ^{4} ) ) ^{6}} d x

x \to 0 lim (\frac{sin ( 4 x )}{9})

What is the integral of-8xsin(4x^2+1) ?
The integral of-8xsin(4x^2+1) is cos(4x^2+1)+C