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

The integral of x/2 cos^2(x) is 1/16 (2x^2+2xsin(2x)+cos(2x))+C

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integral of x/2 cos^2(x)

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

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

Solution steps

\int \frac{x}{2} cos^{2} (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 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

= \frac{1}{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))

= \frac{1}{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

\frac{1}{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))) : \frac{1}{16} (2 x^{2} + 2 x sin (2 x) + cos (2 x))

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

Add a constant to the solution

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

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What is the integral of x/2 cos^2(x) ?
The integral of x/2 cos^2(x) is 1/16 (2x^2+2xsin(2x)+cos(2x))+C