What is the integral of 8x^3sin(1-x^4) ?

The integral of 8x^3sin(1-x^4) is 2cos(1-x^4)+C

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integral of 8x^3sin(1-x^4)

\int 8 x^{3} sin (1 - x^{4}) d x

2 cos (1 - x^{4}) + C

Solution steps

\int 8 x^{3} sin (1 - x^{4}) d x

Take the constant out:

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

= 8 \cdot \int x^{3} sin (1 - x^{4}) d x

Apply u-substitution

: \int - \frac{sin ( u )}{4} d u

= 8 \cdot \int - \frac{sin ( u )}{4} d u

Take the constant out:

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

= 8 (- \frac{1}{4} \cdot \int sin (u) d u)

Use the common integral:

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

= 8 (- \frac{1}{4} (- cos (u)))

Substitute back

u = 1 - x^{4}

= 8 (- \frac{1}{4} (- cos (1 - x^{4})))

Simplify

8 (- \frac{1}{4} (- cos (1 - x^{4}))) : 2 cos (1 - x^{4})

= 2 cos (1 - x^{4})

Add a constant to the solution

= 2 cos (1 - x^{4}) + C

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What is the integral of 8x^3sin(1-x^4) ?
The integral of 8x^3sin(1-x^4) is 2cos(1-x^4)+C