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

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

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integral of 2xcos^2(x)

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

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

Solution steps

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

Take the constant out:

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

= 2 \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

= 2 (\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))

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

Simplify

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

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

Add a constant to the solution

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

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

x \to 0 + lim (e^{\frac{2}{x}})

\int 2 y sin (x) d y

\frac{d y}{d x} = (1 + x) (1 + y)

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

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