1.

1+ COSXdx =... +sinx coSX​

Answer»

ANSWER:

∫1−(1−2sin22x)x−sinxdx=∫2sin22xx−sinxdx

=21∫sin22xx−21∫sin22x2sin2xcos2xdx

=21∫x⋅cosec22x−∫cot2xdx

=21[x⋅∫cosec22x−∫{DXD<



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