What is the integral of 9xcos(8x) ?

The integral of 9xcos(8x) is 9/64 (8xsin(8x)+cos(8x))+C

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integral of 9xcos(8x)

\int 9 x cos (8 x) d x

\frac{9}{64} (8 x sin (8 x) + cos (8 x)) + C

Solution steps

\int 9 x cos (8 x) d x

Take the constant out:

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

= 9 \cdot \int x cos (8 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{64} d u

= 9 \cdot \int \frac{u cos ( u )}{64} d u

Take the constant out:

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

= 9 \cdot \frac{1}{64} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 9 \cdot \frac{1}{64} (u sin (u) - \int sin (u) d u)

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

= 9 \cdot \frac{1}{64} (u sin (u) - (- cos (u)))

Substitute back

u = 8 x

= 9 \cdot \frac{1}{64} (8 x sin (8 x) - (- cos (8 x)))

Simplify

9 \cdot \frac{1}{64} (8 x sin (8 x) - (- cos (8 x))) : \frac{9}{64} (8 x sin (8 x) + cos (8 x))

= \frac{9}{64} (8 x sin (8 x) + cos (8 x))

Add a constant to the solution

= \frac{9}{64} (8 x sin (8 x) + cos (8 x)) + C

y^{^{''}} + 4 y^{^{'}} = 0

\frac{\partial}{\partial y} (x z^{2})

\frac{d ^{3}}{d x ^{3}} (\frac{4}{x ^{2}})

\int_{- 1}^{2} (y + 2 - y^{2}) d y

\frac{e ^{\frac{- π}{2}}}{s ^{2} + 9}

What is the integral of 9xcos(8x) ?
The integral of 9xcos(8x) is 9/64 (8xsin(8x)+cos(8x))+C