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

The integral of xcos^2(4x) is 1/128 (32x^2+8xsin(8x)+cos(8x))+C

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integral of xcos^2(4x)

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

\frac{1}{128} (32 x^{2} + 8 x sin (8 x) + cos (8 x)) + C

Solution steps

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

Rewrite using trig identities

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

Take the constant out:

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

= \frac{1}{2} \cdot \int x (1 + cos (8 x)) d x

Apply Integration By Parts

: x (x + \frac{1}{8} sin (8 x)) - \int x + \frac{1}{8} sin (8 x) d x

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

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

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

Simplify

\frac{1}{2} (x (x + \frac{1}{8} sin (8 x)) - (\frac{x ^{2}}{2} - \frac{1}{64} cos (8 x))) : \frac{1}{128} (32 x^{2} + 8 x sin (8 x) + cos (8 x))

= \frac{1}{128} (32 x^{2} + 8 x sin (8 x) + cos (8 x))

Add a constant to the solution

= \frac{1}{128} (32 x^{2} + 8 x sin (8 x) + cos (8 x)) + C

\frac{d}{d x} (\frac{1}{( x ^{2} - 1 )})

- 45 y^{^{''}} + 35 y^{^{'}} = 0, y (2) = 2, y^{^{'}} (- 1) = - 1

(x + e^{y}) d y - d x = 0

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

\frac{d}{d x} (x + 9)

What is the integral of xcos^2(4x) ?
The integral of xcos^2(4x) is 1/128 (32x^2+8xsin(8x)+cos(8x))+C