What is the integral of (x^3+1)cos(x^4+4x) ?

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

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integral of (x^3+1)cos(x^4+4x)

\int (x^{3} + 1) cos (x^{4} + 4 x) d x

\frac{1}{4} sin (x^{4} + 4 x) + C

Solution steps

\int (x^{3} + 1) cos (x^{4} + 4 x) d x

Apply u-substitution

= \int \frac{cos ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{4} sin (u)

Substitute back

u = x^{4} + 4 x

= \frac{1}{4} sin (x^{4} + 4 x)

Add a constant to the solution

= \frac{1}{4} sin (x^{4} + 4 x) + C

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