What is the integral of θsin(θ)cos(θ) ?

The integral of θsin(θ)cos(θ) is 1/2 (-1/2 θcos(2θ)+1/4 sin(2θ))+C

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integral of θsin(θ)cos(θ)

\int θ sin (θ) cos (θ) d θ

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

Solution steps

\int θ sin (θ) cos (θ) d θ

Rewrite using trig identities

= \int \frac{1}{2} θ sin (2 θ) d θ

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((7 x + 8 y)^{3})

x \to \infty lim (\frac{x ^{2}}{x ^{4} + 2})

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

2 t x d x + (t^{2} - x^{2}) d t = 0

y = \frac{9}{x}

What is the integral of θsin(θ)cos(θ) ?
The integral of θsin(θ)cos(θ) is 1/2 (-1/2 θcos(2θ)+1/4 sin(2θ))+C