MathematicsMedium210×since 2002Q3408If the real part of the complex number (1−cosθ+2isinθ)−1{(1 - \cos \theta + 2i\sin \theta )^{ - 1}}(1−cosθ+2isinθ)−1 is 15{1 \over 5}51 for θ∈(0,π)\theta \in (0,\pi )θ∈(0,π), then the value of the integral ∫0θsinxdx\int_0^\theta {\sin x} dx∫0θsinxdx is equal to:A1B2C−-−1D0Check answerSkip