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Q.Integrate: ∫dxsin⁡2xcos⁡2x\displaystyle\int \dfrac{dx}{\sin^2x\cos^2x}

Tripura TbseHigher Secondary (+2 Stage) Examination 2025Subjective· 2mImportance★★★★★
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Write 1=sin⁡2x+cos⁡2x1=\sin^2x+\cos^2x in the numerator to split the integrand into two standard integrable pieces.

1sin⁡2xcos⁡2x=sin⁡2x+cos⁡2xsin⁡2xcos⁡2x=sin⁡2xsin⁡2xcos⁡2x+cos⁡2xsin⁡2xcos⁡2x=1cos⁡2x+1sin⁡2x=sec⁡2x+csc⁡2x.\dfrac{1}{\sin^2x\cos^2x}=\dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}=\dfrac{\sin^2x}{\sin^2x\cos^2x}+\dfrac{\cos^2x}{\sin^2x\cos^2x}=\dfrac{1}{\cos^2x}+\dfrac{1}{\sin^2x}=\sec^2x+\csc^2x.

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