∫ (\sqrt{tanx}+\sqrt{cottx}) dx =

Correct Answer is : √2sin−1 (sin x - cos x) + x

\sqrt2 \sin^{-1} (sin x + cos x) + x

\sqrt2 \cos^{-1} (sin x + cos x) + x

\sqrt2 \cos^{-1} (sin x - cos x) + x

\sqrt2 \sin^{-1} (sin x - cos x) + x

TS EAMCET 5 May 2018 Shift 1 Solved Paper

Section: Physics

Question : 111 of 160

Marks: +1, -0

∫ (√tanx+√cottx) dx =

Go to Question:

Prev Question

Next Question