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a
=12(b+a)logacos2x+bsin2x+C
b
=12(b−a)logacos2-bsin2x+C
c
=1(b−a)logacos2x+bsin2x+C
d
=12(b−a)logacos2x+bsin2x+C
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Correct option is D
I=∫tanxa+btan2xdx=∫sinxcosxa+bsin2xcos2xdx=∫sinxcosxacos2x+bsin2xdx=12∫sin2xacos2x+bsin2xdx Put acos2x+bsin2x=t∴ (b−a)sin2xdx=dt∴ l=12(b−a)∫dtt =12(b−a)loge|t|+C =12(b−a)logacos2x+bsin2x+CSimilar Questions
is equal to
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