What is the integral of 8xcos(4x^2-5) ?

The integral of 8xcos(4x^2-5) is sin(4x^2-5)+C

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integral of 8xcos(4x^2-5)

\int 8 x cos (4 x^{2} - 5) d x

sin (4 x^{2} - 5) + C

Solution steps

\int 8 x cos (4 x^{2} - 5) d x

Take the constant out:

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

= 8 \cdot \int x cos (4 x^{2} - 5) d x

Apply u-substitution

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

= 8 \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

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

Use the common integral:

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

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

Substitute back

u = 4 x^{2} - 5

= 8 \cdot \frac{1}{8} sin (4 x^{2} - 5)

Simplify

8 \cdot \frac{1}{8} sin (4 x^{2} - 5) : sin (4 x^{2} - 5)

= sin (4 x^{2} - 5)

Add a constant to the solution

= sin (4 x^{2} - 5) + C

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What is the integral of 8xcos(4x^2-5) ?
The integral of 8xcos(4x^2-5) is sin(4x^2-5)+C