What is the integral of x^2cos(1/8 x) ?

The integral of x^2cos(1/8 x) is 8x^2sin(1/8 x)-16(-8xcos(1/8 x)+64sin(1/8 x))+C

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integral of x^2cos(1/8 x)

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

8 x^{2} sin (\frac{1}{8} x) - 16 (- 8 x cos (\frac{1}{8} x) + 64 sin (\frac{1}{8} x)) + C

Solution steps

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

Apply Integration By Parts

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

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

\int 16 x sin (\frac{1}{8} x) d x = 16 (- 8 x cos (\frac{1}{8} x) + 64 sin (\frac{1}{8} x))

= 8 x^{2} sin (\frac{1}{8} x) - 16 (- 8 x cos (\frac{1}{8} x) + 64 sin (\frac{1}{8} x))

Add a constant to the solution

= 8 x^{2} sin (\frac{1}{8} x) - 16 (- 8 x cos (\frac{1}{8} x) + 64 sin (\frac{1}{8} x)) + C

\frac{\partial}{\partial x} (3 sin (x) sin (y))

\int \frac{4 - x}{4 + x} d x

\int (y^{\frac{2}{3}} - 4 y^{- \frac{1}{5}} + 4) d y

\frac{d}{d x} (ln (\frac{x + 1}{x - 3}))

f (x) = \frac{5}{x ^{3}}

What is the integral of x^2cos(1/8 x) ?
The integral of x^2cos(1/8 x) is 8x^2sin(1/8 x)-16(-8xcos(1/8 x)+64sin(1/8 x))+C