What is the integral of 5xsin(x)cos(x) ?

The integral of 5xsin(x)cos(x) is 5/2 (-1/2 xcos(2x)+1/4 sin(2x))+C

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integral of 5xsin(x)cos(x)

\int 5 x sin (x) cos (x) d x

\frac{5}{2} (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) + C

Solution steps

\int 5 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

= 5 \cdot \int x sin (x) cos (x) d x

Rewrite using trig identities

= 5 \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

= 5 \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

= 5 \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)

= 5 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - (- \frac{1}{4} sin (2 x)))

Simplify

5 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - (- \frac{1}{4} sin (2 x))) : \frac{5}{2} (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))

= \frac{5}{2} (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))

Add a constant to the solution

= \frac{5}{2} (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) + C

\frac{d}{d θ} (r (θ) cos (θ))

\int x^{- 2} (4 x^{- 1} + 3)^{\frac{3}{4}} d x

1 0^{^{'}}

2 x y^{2} + x^{2} \frac{d y}{d x} = y^{2}

\int e^{- θ} cos (3 θ) d θ

What is the integral of 5xsin(x)cos(x) ?
The integral of 5xsin(x)cos(x) is 5/2 (-1/2 xcos(2x)+1/4 sin(2x))+C