What is the integral of 8xcos^2(x) ?

The integral of 8xcos^2(x) is 8(1/2 x(x+1/2 sin(2x))-1/2 ((x^2)/2-1/4 cos(2x)))+C

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integral of 8xcos^2(x)

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

8 (\frac{1}{2} x (x + \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x))) + C

Solution steps

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

Take the constant out:

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

= 8 \cdot \int x cos^{2} (x) d x

Apply Integration By Parts

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

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

\int \frac{1}{2} (x + \frac{1}{2} sin (2 x)) d x = \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x))

= 8 (\frac{1}{2} x (x + \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x)))

Add a constant to the solution

= 8 (\frac{1}{2} x (x + \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x))) + C

y^{^{'}} + sin (t) y = - sin (t)

\frac{\partial}{\partial x} (arcsin (10 x yz))

y = \frac{1 + x}{1 + e ^{x}}, (0, \frac{1}{2})

x \to 0 + lim (3 x ln (x))

3 x ln (15 x)

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