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Q.

If ∫tan4⁡xdx=Ktan3⁡x+Ltan⁡x+f(x), then

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a

K=2/3

b

L=−2

c

f(x)=x+C

d

K=4/3

answer is C.

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Detailed Solution

∫tan4⁡xdx=∫tan2⁡xsec2⁡x−1dx=tan3⁡x3−∫sec2⁡x−1dx=tan2⁡x3−tan⁡x+x+C.
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