\(\int\frac{cosx}{\sqrt{4-sin^2x}}\) = ?∫cosx/√{4 - sin2x}A.sin-1

1.	\(\int\frac{cosx}{\sqrt{4-sin^2x}}\) = ?∫cosx/√{4 - sin2x}A.sin-1x/2 + CB.sin-1(1/2cosx) + CC.sin-1(2sinx) + CD. sin-1(1/2sinx) + C
Answer» Put sin x = t ⇒ cos x dx = dt ∴ The given equation becomes \(\int\frac{dt}{\sqrt{4-t^2}}\) = sin^-1 t/2 + C But t = sin x = sin^-1\((\frac{sinx}2)\) + C

\(\int\frac{cosx}{\sqrt{4-sin^2x}}\) = ?∫cosx/√{4 - sin2x}A.sin-1x/2 + CB.sin-1(1/2cosx) + CC.sin-1(2sinx) + CD. sin-1(1/2sinx) + C

Answer»

Put sin x = t

⇒ cos x dx = dt

∴ The given equation becomes

\(\int\frac{dt}{\sqrt{4-t^2}}\)

= sin^-1 t/2 + C

But t = sin x

= sin^-1\((\frac{sinx}2)\) + C