What is the integral of 16xsin(x)cos(x) ?

The integral of 16xsin(x)cos(x) is 8(-1/2 xcos(2x)+1/4 sin(2x))+C

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integral of 16xsin(x)cos(x)

\int 16 x sin (x) cos (x) d x

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

Solution steps

\int 16 x sin (x) cos (x) d x

Take the constant out:

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

= 16 \cdot \int x sin (x) cos (x) d x

Rewrite using trig identities

= 16 \cdot \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

= 16 \cdot \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

= 16 \cdot \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)

= 16 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - (- \frac{1}{4} sin (2 x)))

Simplify

16 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - (- \frac{1}{4} sin (2 x))) : 8 (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))

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

Add a constant to the solution

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

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What is the integral of 16xsin(x)cos(x) ?
The integral of 16xsin(x)cos(x) is 8(-1/2 xcos(2x)+1/4 sin(2x))+C