What is the integral from x to 1 of 2θsin(θ) ?

The integral from x to 1 of 2θsin(θ) is 2(-cos(1)+sin(1)+xcos(x)-sin(x))

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integral from x to 1 of 2θsin(θ)

\int_{x}^{1} 2 θ sin (θ) d θ

2 (- cos (1) + sin (1) + x cos (x) - sin (x))

Solution steps

\int_{x}^{1} 2 θ sin (θ) d θ

Take the constant out:

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

= 2 \cdot \int_{x}^{1} θ sin (θ) d θ

Apply Integration By Parts

= 2 [- θ cos (θ) - \int - cos (θ) d θ]_{x}^{1}

\int - cos (θ) d θ = - sin (θ)

= 2 [- θ cos (θ) - (- sin (θ))]_{x}^{1}

Simplify

= 2 [- θ cos (θ) + sin (θ)]_{x}^{1}

Compute the boundaries:

- cos (1) + sin (1) + x cos (x) - sin (x)

= 2 (- cos (1) + sin (1) + x cos (x) - sin (x))

What is the integral from x to 1 of 2θsin(θ) ?
The integral from x to 1 of 2θsin(θ) is 2(-cos(1)+sin(1)+xcos(x)-sin(x))