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B
1√5tan−1(tanx√5)+c
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C
12√5tan−1(2tanx√5)+c
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D
None of these
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Solution
The correct option is C12√5tan−1(2tanx√5)+c ∫dx4sin2x+5cos2x=∫sec2xdx4tan2x+5=14∫sec2xdxtan2x+54 Put tanx=t⇒sec2xdx=dt, then it reduces to 14∫dtt2+(√52)2=24√5tan−1(2t√5)+c=12√5tan−1(2tanx√5)+c