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Question:

Let n≥2 be a natural number and 0<θ<π/2. Then ∫(sinπθ−sinθ)1/(πcosθsinπ+1θ)dθ is equal to: (Where C is a constant of integration)

nn2(1+1sinπθ)π+1π+C

1n2+1(1sinπθ)π+1π+C

1n(1sinπθ)π+1π+C

nn2(1sinπ+1θ)π+1π+C

Solution:

∫(sinπθ−sinθ)1/(πcosθsinπ+1θ)dθ=∫sinθ(1−sinπθ)1/(πcosθsinπ+1θ)dθ
Put 1−sinπθ=t
So (−nπ)sinπθcosθdθ=dt
Now 1n∫(t)1/πdt=1/(nπ)(t)π+1π+C=1/(nπ)(1−sinπθ)π+1π+C
1n(1−sinπθ)π+1π+C