What is the integral of-64x^3sin(4x^4+3) ?

The integral of-64x^3sin(4x^4+3) is 4cos(4x^4+3)+C

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integral of-64x^3sin(4x^4+3)

\int - 64 x^{3} sin (4 x^{4} + 3) d x

4 cos (4 x^{4} + 3) + C

Solution steps

\int - 64 x^{3} sin (4 x^{4} + 3) d x

Take the constant out:

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

= - 64 \cdot \int x^{3} sin (4 x^{4} + 3) d x

Apply u-substitution

: \int \frac{sin ( u )}{16} d u

= - 64 \cdot \int \frac{sin ( u )}{16} d u

Take the constant out:

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

= - 64 \cdot \frac{1}{16} \cdot \int sin (u) d u

Use the common integral:

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

= - 64 \cdot \frac{1}{16} (- cos (u))

Substitute back

u = 4 x^{4} + 3

= - 64 \cdot \frac{1}{16} (- cos (4 x^{4} + 3))

Simplify

- 64 \cdot \frac{1}{16} (- cos (4 x^{4} + 3)) : 4 cos (4 x^{4} + 3)

= 4 cos (4 x^{4} + 3)

Add a constant to the solution

= 4 cos (4 x^{4} + 3) + C

h \to 0 lim (\frac{( 1 + h ) ^{7} - 1}{h})

f (x) = \frac{4}{3} x - 7

\frac{d ^{3}}{d x ^{3}} (\frac{1}{x ^{2}})

\frac{\partial}{\partial x} ((1 - x)^{2})

\frac{d}{d t} (\frac{c e ^{t}}{1 + c e ^{t}})

What is the integral of-64x^3sin(4x^4+3) ?
The integral of-64x^3sin(4x^4+3) is 4cos(4x^4+3)+C