What is the integral of [1+sin^2(θ)csc(θ)] ?

The integral of [1+sin^2(θ)csc(θ)] is θ-cos(θ)+C

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integral of [1+sin^2(θ)csc(θ)]

\int [1 + sin^{2} (θ) csc (θ)] d θ

θ - cos (θ) + C

Solution steps

\int 1 + sin^{2} (θ) csc (θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d θ + \int sin^{2} (θ) csc (θ) d θ

\int 1 d θ = θ

\int sin^{2} (θ) csc (θ) d θ = - cos (θ)

= θ - cos (θ)

Add a constant to the solution

= θ - cos (θ) + C

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What is the integral of [1+sin^2(θ)csc(θ)] ?
The integral of [1+sin^2(θ)csc(θ)] is θ-cos(θ)+C