What is the integral of xcos(8x^2) ?

The integral of xcos(8x^2) is 1/16 sin(8x^2)+C

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integral of xcos(8x^2)

\int x cos (8 x^{2}) d x

\frac{1}{16} sin (8 x^{2}) + C

Solution steps

\int x cos (8 x^{2}) d x

Apply u-substitution

: \int \frac{cos ( u )}{16} d u

= \int \frac{cos ( u )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{16} sin (u)

Substitute back

u = 8 x^{2}

= \frac{1}{16} sin (8 x^{2})

Add a constant to the solution

= \frac{1}{16} sin (8 x^{2}) + C

f (x) = x^{2} - 2 x + 3, at x = 2

x \to 0 lim (Q (x))

\int \frac{7}{3 p - 4} d p

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

\int \frac{16 x ^{2}}{( x - 15 ) ( x + 5 ) ^{2}} d x

What is the integral of xcos(8x^2) ?
The integral of xcos(8x^2) is 1/16 sin(8x^2)+C