If \int\frac{acosx-2sinx}{bsinx+5cosx}dx=7/{41}x+\frac{22}{41}\log|bsinx+5cosx|+C,(a>0,b>0) then \int\frac{dx}{b+acosx}=

Correct Answer is : 2√7Tan−1(tanx2√7)+C

2/3\log({3tanx/2+4-\sqrt{3}/{3tanx/2+4+\sqrt{3})+C

2/\sqrt{7}Tan^{-1}({tanx/2}/\sqrt{7})+C

2/\sqrt{7}\log(\sqrt{7}-tanx/2}/\sqrt{7}+tanx/2})+C

2Sinh^{-1}(\frac{2tanx/2}{1+\tan^2x/2})+C

TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

If ∫

acosx−2sinx

bsinx+5cosx

dx=

log|bsinx+5cosx|+C,(a>0,b>0)
then ∫

b+acosx

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