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Q.

∫1sin2x cos2x−sin4x dx=

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a

− cos−1(cotx)+C

b

− cosh−1(cotx)+C

c

cosh−1(cotx)+C

d

cos−1(cotx)+C

answer is B.

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Detailed Solution

∫1sin2xcot2x−1dx              =∫cosec2xcot2x−1dx  cotx=t⇒−cosec2xdx=dt             =−∫1t2−1dt             =−cosh−1(cotx)+C
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