What is the integral of 60x^9sin(3x^{10}) ?

The integral of 60x^9sin(3x^{10}) is -2cos(3x^{10})+C

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integral of 60x^9sin(3x^{10})

\int 60 x^{9} sin (3 x^{10}) d x

- 2 cos (3 x^{10}) + C

Solution steps

\int 60 x^{9} sin (3 x^{10}) d x

Take the constant out:

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

= 60 \cdot \int x^{9} sin (3 x^{10}) d x

Apply u-substitution

: \int \frac{sin ( u )}{30} d u

= 60 \cdot \int \frac{sin ( u )}{30} d u

Take the constant out:

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

= 60 \cdot \frac{1}{30} \cdot \int sin (u) d u

Use the common integral:

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

= 60 \cdot \frac{1}{30} (- cos (u))

Substitute back

u = 3 x^{10}

= 60 \cdot \frac{1}{30} (- cos (3 x^{10}))

Simplify

60 \cdot \frac{1}{30} (- cos (3 x^{10})) : - 2 cos (3 x^{10})

= - 2 cos (3 x^{10})

Add a constant to the solution

= - 2 cos (3 x^{10}) + C

3 x + 1, at x = 5

\frac{\partial}{\partial x} (e^{x} - 1)

\frac{d}{d x} (\frac{1}{5 x ^{\frac{4}{5}}})

\int w d w

\int x^{3} + 2 x d x

What is the integral of 60x^9sin(3x^{10}) ?
The integral of 60x^9sin(3x^{10}) is -2cos(3x^{10})+C