What is the integral of (90x^2sin(2+6x^3)) ?

The integral of (90x^2sin(2+6x^3)) is -5cos(2+6x^3)+C

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integral of (90x^2sin(2+6x^3))

\int (90 x^{2} sin (2 + 6 x^{3})) d x

- 5 cos (2 + 6 x^{3}) + C

Solution steps

\int 90 x^{2} sin (2 + 6 x^{3}) d x

Take the constant out:

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

= 90 \cdot \int x^{2} sin (2 + 6 x^{3}) d x

Apply u-substitution

: \int \frac{u sin ( 2 + 6 u ^{\frac{3}{2}} )}{2 x} d u

= 90 \cdot \int \frac{sin ( 2 + 6 u )}{3} d u

Take the constant out:

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

= 90 \cdot \frac{1}{3} \cdot \int sin (2 + 6 u) d u

Apply u-substitution

: \int sin (v) \frac{1}{6} d v

= 90 \cdot \frac{1}{3} \cdot \int sin (v) \frac{1}{6} d v

Take the constant out:

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

= 90 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \int sin (v) d v

Use the common integral:

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

= 90 \cdot \frac{1}{3} \cdot \frac{1}{6} (- cos (v))

Substitute back

= 90 \cdot \frac{1}{3} \cdot \frac{1}{6} (- cos (2 + 6 x^{3}))

Simplify

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

= - 5 cos (2 + 6 x^{3})

Add a constant to the solution

= - 5 cos (2 + 6 x^{3}) + C

\frac{d}{d x} (\frac{3 x + 7}{5 + 2 x})

\frac{d}{d x} ((ln (x + 1))^{2} - e^{x^{2}})

\int ∣ 4 y + 1 ∣ d y

y^{^{'}} + 7 y = t + e^{- 6 t}

2 x d x + d y = 0

What is the integral of (90x^2sin(2+6x^3)) ?
The integral of (90x^2sin(2+6x^3)) is -5cos(2+6x^3)+C