What is the integral from 0 to 2sin(θ) of r^2sin(θ) ?

The integral from 0 to 2sin(θ) of r^2sin(θ) is r^2(-cos(2sin(θ))+1)

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integral from 0 to 2sin(θ) of r^2sin(θ)

\int_{0}^{2 s i n (θ)} r^{2} sin (θ) d θ

r^{2} (- cos (2 sin (θ)) + 1)

Solution steps

\int_{0}^{2 s i n (θ)} r^{2} sin (θ) d θ

Take the constant out:

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

= r^{2} \cdot \int_{0}^{2 s i n (θ)} sin (θ) d θ

Use the common integral:

\int sin (θ) d θ = - cos (θ)

= r^{2} [- cos (θ)]_{0}^{2 s i n (θ)}

Compute the boundaries:

- cos (2 sin (θ)) + 1

= r^{2} (- cos (2 sin (θ)) + 1)

\int_{- 4}^{1} - x^{2} - 3 x + 4 d x

\int_{0}^{1} \frac{sin ^{2} ( x )}{x ^{2}} d x

\int_{- 2 π}^{2 π} x^{2} cos (2 x) d x

\int_{1}^{2} 4 d x

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

What is the integral from 0 to 2sin(θ) of r^2sin(θ) ?
The integral from 0 to 2sin(θ) of r^2sin(θ) is r^2(-cos(2sin(θ))+1)