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$\int \frac{\sin 2 x}{\sin 5 x \sin 3 x} d x =$

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logsinx−logcosx+c

12logsin2x+13logsin3x+c

13logsin2x+15logsin3x+c

13logsin3x−15logsin5x+c

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detailed solution

Correct option is D

∫sin2xsin5xsin3xdx=∫sin⁡(5x−3x)sin⁡5x⋅sin⁡3xdx=∫sin⁡5xcos⁡3x−cos⁡5xsin⁡3xsin⁡5x⋅sin⁡3xdx=∫cot⁡3xdx−∫cot⁡5xdx=13log⁡|sin⁡3x|-15|sin⁡5x|+c

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∫sin2xsin5x sin3xdx=

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$\int \frac{\sin 2 x}{\sin 5 x \sin 3 x} d x =$