What is the integral of x^2sin(4x^3+4^3) ?

The integral of x^2sin(4x^3+4^3) is -1/12 cos(4x^3+64)+C

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integral of x^2sin(4x^3+4^3)

\int x^{2} sin (4 x^{3} + 4^{3}) d x

- \frac{1}{12} cos (4 x^{3} + 64) + C

Solution steps

\int x^{2} sin (4 x^{3} + 4^{3}) d x

Simplify

= \int x^{2} sin (4 x^{3} + 64) d x

Apply u-substitution

: \int \frac{u sin ( 4 u ^{\frac{3}{2}} + 64 )}{2 x} d u

= \int \frac{sin ( 4 u + 64 )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sin (4 u + 64) d u

Apply u-substitution

: \int sin (v) \frac{1}{4} d v

= \frac{1}{3} \cdot \int sin (v) \frac{1}{4} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{4} \cdot \int sin (v) d v

Use the common integral:

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

= \frac{1}{3} \cdot \frac{1}{4} (- cos (v))

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{1}{4} (- cos (4 x^{3} + 64)) : - \frac{1}{12} cos (4 x^{3} + 64)

= - \frac{1}{12} cos (4 x^{3} + 64)

Add a constant to the solution

= - \frac{1}{12} cos (4 x^{3} + 64) + C

\int \frac{( cos ^{2} ( ln ( x ) ) )}{x} d x

\int x^{2} sin (ln (x)) d x

\int \frac{x ^{2}}{1 + x - x ^{2}} d x

\int \frac{x ^{3} + 1}{x ^{4} + 4 x} d x

\int \frac{4 x ^{2}}{1 + x ^{2}} d x

What is the integral of x^2sin(4x^3+4^3) ?
The integral of x^2sin(4x^3+4^3) is -1/12 cos(4x^3+64)+C