What is the integral of 6xsin(3x^2+5) ?

The integral of 6xsin(3x^2+5) is -cos(3x^2+5)+C

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integral of 6xsin(3x^2+5)

\int 6 x sin (3 x^{2} + 5) d x

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

Solution steps

\int 6 x sin (3 x^{2} + 5) d x

Take the constant out:

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

= 6 \cdot \int x sin (3 x^{2} + 5) d x

Apply u-substitution

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

= 6 \cdot \int \frac{sin ( 3 u + 5 )}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int sin (3 u + 5) d u

Apply u-substitution

: \int sin (v) \frac{1}{3} d v

= 6 \cdot \frac{1}{2} \cdot \int sin (v) \frac{1}{3} d v

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \int sin (v) d v

Use the common integral:

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

= 6 \cdot \frac{1}{2} \cdot \frac{1}{3} (- cos (v))

Substitute back

= 6 \cdot \frac{1}{2} \cdot \frac{1}{3} (- cos (3 x^{2} + 5))

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 6xsin(3x^2+5) ?
The integral of 6xsin(3x^2+5) is -cos(3x^2+5)+C