What is the integral of 16xcos(8x^2-1) ?

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

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integral of 16xcos(8x^2-1)

\int 16 x cos (8 x^{2} - 1) d x

sin (8 x^{2} - 1) + C

Solution steps

\int 16 x cos (8 x^{2} - 1) d x

Take the constant out:

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

= 16 \cdot \int x cos (8 x^{2} - 1) d x

Apply u-substitution

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

= 16 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = 8 x^{2} - 1

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

Simplify

16 \cdot \frac{1}{16} sin (8 x^{2} - 1) : sin (8 x^{2} - 1)

= sin (8 x^{2} - 1)

Add a constant to the solution

= sin (8 x^{2} - 1) + C

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