$sin^2x cos2x dx

[Riazy]

$sin^2x cos2x dx

Right this is the problem,
see my scanned paper.

 

[mmm4444bot]

Your result in green is correct. Is this confirmation the reason for your post? I don't see any question.

 

[Riazy]

They right answer should be the green, but I didnt get there, if someone could help me get there i would really b grateful

 

[Deleted member 4993]

$\displaystyle \int sin^2(x).cos(2x) dx$

we know:

$\displaystyle sin^2(x)\ = \ \frac{1-cos(2x)}{2}$

and

$\displaystyle cos^2(2x)\ = \ \frac{1+ cos(4x)}{2}$

then

$\displaystyle \int sin^2(x).cos(2x) dx$

$\displaystyle = \ \int \frac{1-cos(2x)}{2} \cdot cos(2x) dx$

$\displaystyle = \ \frac{1}{2}\int [cos(2x) \ - \ cos^2(2x)] dx$

$\displaystyle = \ \frac{1}{4}sin(2x) \ - \frac{1}{2}\int [cos^2(2x)] dx$

$\displaystyle = \ \frac{1}{4}sin(2x) \ - \frac{1}{2}\int \frac{1+ cos(4x)}{2} dx$

Continue and finish it.....