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If dxsin4x+cos4x=12tan1f(x)+C then

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a
f(x)=tan⁡x−cot⁡x
b
f(π/4)=0
c
f (x) is continuous on R
d
f(x)=12(tan⁡x−cot⁡x)

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detailed solution

Correct option is B

Dividing the numerator and denominator by cos4⁡x, the given integral can be written as ∫sec2⁡x1+tan2⁡xtan4⁡x+1dx=∫1+t21+t4dt(t=tan⁡x)=∫1+1/t2t2+1/t2dt=∫1+1/t2t−1t2+2dt=12tan−1⁡12t−1t+C=12tan−1⁡tan⁡x−cot⁡x2+CThus f(x)=tan⁡x−cot⁡x2 so fπ4=0


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