Question

Compute this intergral

Answer 1

\[\int\limits_{}^{}\sin2\omega \cos2\omega d \omega\]

Answer 2

(2∫sin2ωcos2ωdω)/2
1/2 * ∫sin(4ω)dω
-1/8*cos4ω+c

Answer 3

Put t = sin(2theta) \[\frac{dt}{d{\theta}} = 2 \cos2{\theta} \] \[\frac{d{\theta}}{1} = \frac{dt}{2cos2{\theta}}\]

Answer 4

Then Just Replace the Values

\[ \int \frac{sin2{\theta}cos2{\theta}}{2cos2{\theta}}d{t}\]
\[\frac{1}{2} \times \int\frac{t}{1}dt\]
\[\frac{1}{4}\times t^2 \]
\[t^2 = sin^2 2{\theta}\]

So, the Answer is

\[\frac{sin^22{\theta}}{4}\]

Answer 5

thanks alot!!