∫sinx/√{4cos^2 x - 1} = ?

Question

\(\int\frac{sinx}{\sqrt{4cos^2x-1}}\) = ?

∫sinx/√{4cos²x - 1}

A.-\(\frac{1}2\)log | 2cosx + \(\sqrt{4cos^2x-1}\) + c

B.-\(\frac{1}2\)log | 2cosx + \(\sqrt{4cos^2x-1}\) + c

C.- \(\frac{1}2\)log | 2cosx + \(\sqrt{4cos^2x-1}\) + c

D. none of these

Answer 1

⇒ \(\int\frac{sinxdx}{\sqrt{(2cosx)^2-(1})^2}\)

Put t = 2cos x 

dt = -2sinx dx