Answer:
[tex]\boxed{\frac{1}{12} \sin^6 2x + c}[/tex]
Step-by-step explanation:
[tex]\int\limits {\sin^5 2x \cos 2x} \, dx[/tex]
[tex]= \int\limits {\sin^5 2x ~.~\frac{1}{2} } \, d(\sin 2x)[/tex]
[tex]= \frac{1}{2} \int\limits {\sin^5 2x } \, d(\sin 2x)[/tex]
[tex]= \frac{1}{2} ~.~\frac{1}{6} ~.~\sin^6 2x + c[/tex]
[tex]= \frac{1}{12} \sin^6 2x + c[/tex]
Step-by-step explanation:
sorry, i don't know lol kgi gi yo gi yig iy iy iy i gi y8 y8
