Question

∫√tanxsinxcosxdx=(a)2√tanx(b)2√cotx(c)√cotx(d)√tanx

Answer 1

Answer:

a) 2√Tanx.

Step-by-step explanation:

∫[√tanx /sinxcosx] dx

We know that Tanx = Sinx/Cosx => Sinx = Tanx/Secx.

= ∫[√Tanx / Tanx/Secx * 1/Secx] dx

= ∫ [Sec²x/√Tanx]dx

Let Tanx = u => du = Sec²xdx.

= ∫ du/√u

= 2√u

= 2√Tanx.