For k∊(1,\infty), \int1/{1+kcosx}dx=

Correct Answer is : 1√k2+1log−1(√k+1+√k−1tanx2√k+1−√k−1tanx2)+C

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

1/\sqrt{k^2-1}\log(\sqrt{k+1}+\sqrt{k-1}tanx/2}/\sqrt{k+1}-\sqrt{k-1})+C

1/\sqrt{k^2+1}\log^{-1}(\sqrt{k+1}+\sqrt{k-1}tanx/2}/\sqrt{k+1}-\sqrt{k-1}tanx/2})+C

1/\sqrt{k^2-1}\Tan{-1}(\sqrt{k+1}cosx/2+\sqrt{k-1}sinx/2}/\sqrt{k+1}cosx/2-\sqrt{k-1}sinx/2})+C

TS EAMCET 10-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

Question : 72 of 160

Marks: +1, -0

For k∊(1,∞),∫

1+kcosx

dx=

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