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a
12logtanx2+π6+c
b
12logtanx2+π12+c
c
12logtanx2+c
d
None of these
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Correct option is B
Let 3=rsinθ and 1=rcosθ Then r=(3)2+12=2 and tanθ=31 or θ=π3 ∴ ∫13sinx+cosxdx =∫1rsinθsinx+rcosθcosxdx =1r∫1cos(x−θ)dx=1r∫sec(x−θ)dx =1rlogtanπ4+x2−θ2+c =12logtanπ4+x2−π6+c =12logtanx2+π12+cSimilar Questions
is equal to
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