What is the integral of 20xsin(5x^2-3) ?

The integral of 20xsin(5x^2-3) is -2cos(5x^2-3)+C

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integral of 20xsin(5x^2-3)

\int 20 x sin (5 x^{2} - 3) d x

- 2 cos (5 x^{2} - 3) + C

Solution steps

\int 20 x sin (5 x^{2} - 3) d x

Take the constant out:

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

= 20 \cdot \int x sin (5 x^{2} - 3) d x

Apply u-substitution

: \int \frac{sin ( 5 u - 3 )}{2} d u

= 20 \cdot \int \frac{sin ( 5 u - 3 )}{2} d u

Take the constant out:

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

= 20 \cdot \frac{1}{2} \cdot \int sin (5 u - 3) d u

Apply u-substitution

: \int sin (v) \frac{1}{5} d v

= 20 \cdot \frac{1}{2} \cdot \int sin (v) \frac{1}{5} d v

Take the constant out:

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

= 20 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot \int sin (v) d v

Use the common integral:

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

= 20 \cdot \frac{1}{2} \cdot \frac{1}{5} (- cos (v))

Substitute back

= 20 \cdot \frac{1}{2} \cdot \frac{1}{5} (- cos (5 x^{2} - 3))

Simplify

20 \cdot \frac{1}{2} \cdot \frac{1}{5} (- cos (5 x^{2} - 3)) : - 2 cos (5 x^{2} - 3)

= - 2 cos (5 x^{2} - 3)

Add a constant to the solution

= - 2 cos (5 x^{2} - 3) + C

f (x) = \frac{9}{x ^{2}}, at x = - 1

f (x) = 2 - 4 x, at x = - 1

f (x) = n (n^{2} + 3 n - 4)^{7}

\int 56 x e^{4 x^{2}} d x

f (z) = e^{5 \cdot z^{2}}

What is the integral of 20xsin(5x^2-3) ?
The integral of 20xsin(5x^2-3) is -2cos(5x^2-3)+C