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 Evaluate 1sin4x+cos4xdx

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a
122tan−1⁡tan2⁡x−12tan⁡x+C
b
12tan−1⁡tan2⁡x−1tan⁡x+C
c
12tan−1⁡tan2⁡x−12tan⁡x+C
d
None of these

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detailed solution

Correct option is C

I=∫1sin4⁡x+cos4⁡xdx=∫1cos4⁡xsin4⁡x+cos4⁡xcos4⁡xdx=∫sec4⁡xtan4⁡x+1dx=∫sec2⁡xsec2⁡xtan4⁡x+1dx=∫1+tan2⁡x1+tan4⁡xsec2⁡xdx Putting tan⁡x=t and sec2⁡xdx=dt , we get I=∫1+t21+t4dt Divide by t2 ∫1t2+11t2+t2dt=Tan−1t−1t+c             ·  here t−1t=u, (1+1t2)dt=du ∫duu2+2 =12tan−1⁡t2−12t+C=12tan−1⁡tan2⁡x−12tan⁡x+C


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