Question

integrate the function √(sin2x)*cos2x.dx

Answer 1

HELLO DEAR,

Given function is ∫√(sin2x)*cos2x.dx

let sin2x = t

2cos2x = dt/dx

dx = dt/(2cos2x)

So, ∫√(sin2x)*cos2x * (dt/2cos2x)

⇒1/2∫√t.dt

⇒1/2[t^{3/2}/(3/2)] + c

put the value of t in above function

⇒(sin2x)^{3/2}/3 + c

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

$\bf{I=\int{\sqrt{sinx}cosx}\,dx}$

Let sin2x = f(x) ----------(1)

differentiate both sides,

2cos2x = f'(x) -----------(2)

put equations (1) and (2) in I,

$\bf{I=\int{\sqrt{f(x)}\frac{f(x)}{2}}\,dx}$

= $\bf{\frac{1}{2}\int{\sqrt{f(x)}f'(x)}\,dx}$

= $\bf{\frac{1}{2}\frac{f(x)^{1/2+1}}{1/2+1}+C }$

= $\bf{\frac{1}{2}.\frac{2}{3}f(x)^{3/2}+C}$

put f(x) = sin2x

therefore, I = $\bf{\frac{1}{3}sin2x\sqrt{sin2x}+C}$