∫ \sqrt\frac{cos2x}{sinx} dx =

Correct Answer is : 1√2log|√2+√1−tan2x√2−√1−tan2x|−12log|1+√1−tan2x1−√1−tan2x|+c

1/{2\sqrt2} \log |{\sqrt2+\sqrt{1-\tan^2x}/{\sqrt2-\sqrt{1-\tan^2x}| - 1/{2} \log |{1+\sqrt{1-\tan^2x}/{1-\sqrt{1-\tan^2x}| + c

1/{\sqrt2} \log |{\sqrt2+\sqrt{1-\tan^2x}/{\sqrt2-\sqrt{1-\tan^2x}| - 1/{2} \log |{1+\sqrt{1-\tan^2x}/{1-\sqrt{1-\tan^2x}| + c

1/{4\sqrt2} \log |{\sqrt2+\sqrt{1-\tan^2x}/{\sqrt2-\sqrt{1-\tan^2x}| - 1/{2} \log |{1+\sqrt{1-\tan^2x}/{1-\sqrt{1-\tan^2x}| + c

1/{4\sqrt2} \log |{\sqrt2-\sqrt{1-\tan^2x}/{\sqrt2+\sqrt{1-\tan^2x}| - 1/{2\sqrt2} \log |{1-\sqrt{1-\tan^2x}/{1+\sqrt{1-\tan^2x}| + c

TS EAMCET 03 May 2019 Shift 1 Solved Paper

Section: Mathematics

Question : 72 of 160

Marks: +1, -0

∫

√cos2x

sinx

dx =

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