What is the integral of x*sin(x)cos(x) ?

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

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integral of x*sin(x)cos(x)

\int x \cdot sin (x) cos (x) d x

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

Solution steps

\int x sin (x) cos (x) d x

Rewrite using trig identities

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

n = 7 \sum \infty \frac{1}{n ln ( n )}

\int_{0}^{3} x^{2} - 9 d x

\int - e^{- 2 x} d x

\frac{64}{x ^{2} + 16}

f (x) = x^{2} - 4 x, at x = 1

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