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? dx/sinx^4+sinx^2cosx^2+cosx^4
multiplying by (sinx^2-cosx^2)in num. and denom. Both, we get
?[(sinx^2-cosx^2)/sinx^4+sinx^2cosx^2+cosx^4]dx
ð ?[(sinx^2-cosx^2)/(sinx^2-cosx^2)^3]dx
ð ?[cos2x/(cos2x)^3]dx
ð ?[1/(cos2x)^2]dx
ð ?[(sec2x)^2]dx
ð (tan2x)/2
ð ½(tan2x) Ans.
Thnaks & Regards
Rinkoo Gupta
AskIITians Faculty.
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