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∫( sin²x/1+cos2x )dx = 1/2tanx - x/2 + c
Step-by-step explanation:
- = ∫sin²x/(1+cos2x)dx
- = ∫sin²x/2cos²x dx [∵1+cos2x = 2cos²x]
- = 1/2∫tan²xdx
- = 1/2∫ (sec²x -1)dx [∵ tan²x = (sec²x -1 ]
- = 1/2tanx - x/2 + c Answer
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