What is the integral of 4xcos(4x^2) ?

The integral of 4xcos(4x^2) is 1/2 sin(4x^2)+C

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integral of 4xcos(4x^2)

\int 4 x cos (4 x^{2}) d x

\frac{1}{2} sin (4 x^{2}) + C

Solution steps

\int 4 x cos (4 x^{2}) d x

Take the constant out:

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

= 4 \cdot \int x cos (4 x^{2}) d x

Apply u-substitution

: \int \frac{cos ( u )}{8} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 4 x^{2}

= 4 \cdot \frac{1}{8} sin (4 x^{2})

Simplify

4 \cdot \frac{1}{8} sin (4 x^{2}) : \frac{1}{2} sin (4 x^{2})

= \frac{1}{2} sin (4 x^{2})

Add a constant to the solution

= \frac{1}{2} sin (4 x^{2}) + C

\frac{d y}{d x} = y - 1

\frac{d y}{d x} = \frac{y ^{3}}{3 x - 2}, y (1) = - 1

\int_{- \infty}^{5} \frac{1}{x ^{2} + 1} d x

\int 1 + cos (θ) d θ

f (x) = - 7 x^{4} + 11 x^{2} - 4, at x = 1

What is the integral of 4xcos(4x^2) ?
The integral of 4xcos(4x^2) is 1/2 sin(4x^2)+C