What is the integral of cos(4x)x ?

The integral of cos(4x)x is 1/16 (4xsin(4x)+cos(4x))+C

integral of cos(4x)x

\int cos (4 x) x d x

\frac{1}{16} (4 x sin (4 x) + cos (4 x)) + C

Solution steps

\int cos (4 x) x d x

Apply u-substitution

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

Apply Integration By Parts

= \frac{1}{16} (u sin (u) - \int sin (u) d u)

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

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

Substitute back

u = 4 x

= \frac{1}{16} (4 x sin (4 x) - (- cos (4 x)))

Simplify

= \frac{1}{16} (4 x sin (4 x) + cos (4 x))

Add a constant to the solution

= \frac{1}{16} (4 x sin (4 x) + cos (4 x)) + C

\int \frac{4}{( x ^{3} + 4 x )} d x

\int \frac{2 x}{9 + x ^{2}} d x

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

\int 3 x^{2} (x^{3} + 2)^{\frac{3}{2}} d x

\int 3 (x^{2} - 4 x + 4)^{4} d x

What is the integral of cos(4x)x ?
The integral of cos(4x)x is 1/16 (4xsin(4x)+cos(4x))+C