What is the integral of 5xcos(4x) ?

The integral of 5xcos(4x) is 5/16 (4xsin(4x)+cos(4x))+C

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integral of 5xcos(4x)

\int 5 x cos (4 x) d x

\frac{5}{16} (4 x sin (4 x) + cos (4 x)) + C

Solution steps

\int 5 x cos (4 x) d x

Take the constant out:

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

= 5 \cdot \int x cos (4 x) d x

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot \frac{1}{16} \cdot \int u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= 5 \cdot \frac{1}{16} (u sin (u) - \int sin (u) d u)

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

= 5 \cdot \frac{1}{16} (u sin (u) - (- cos (u)))

Substitute back

u = 4 x

= 5 \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x)))

Simplify

5 \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x))) : \frac{5}{16} (4 x sin (4 x) + cos (4 x))

= \frac{5}{16} (4 x sin (4 x) + cos (4 x))

Add a constant to the solution

= \frac{5}{16} (4 x sin (4 x) + cos (4 x)) + C

\frac{d}{d x} (\frac{cos ( x )}{1 - sin ( x )})

\int sin^{3} (cos^{2} (d) x) d x

\int (14 x^{6} y^{- 4} - 12 x^{5} y^{- 3}) d x

e^{a r c t a n (3 x^{2})}

\frac{d ^{3}}{d x ^{3}} (\frac{x}{x + 1})

What is the integral of 5xcos(4x) ?
The integral of 5xcos(4x) is 5/16 (4xsin(4x)+cos(4x))+C