What is the integral of x^3cos(x^4-5) ?

The integral of x^3cos(x^4-5) is 1/4 sin(x^4-5)+C

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integral of x^3cos(x^4-5)

\int x^{3} cos (x^{4} - 5) d x

\frac{1}{4} sin (x^{4} - 5) + C

Solution steps

\int x^{3} cos (x^{4} - 5) d x

Apply u-substitution

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

= \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} - 5

= \frac{1}{4} sin (x^{4} - 5)

Add a constant to the solution

= \frac{1}{4} sin (x^{4} - 5) + C

t \to \infty lim (e^{- t})

2 x^{\frac{1}{3}}

\int (2 x + 1) sin (x) d x

\int \frac{x}{x ^{3} + x ^{2} + x + 1} d x

\frac{d}{d x} (a^{2} x + a^{2} y + b^{2} x + b^{2} y)

What is the integral of x^3cos(x^4-5) ?
The integral of x^3cos(x^4-5) is 1/4 sin(x^4-5)+C