$sin^2x cos2x dx
- Thread starter Riazy
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mmm4444bot
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Your result in green is correct. Is this confirmation the reason for your post? I don't see any question.
D
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\(\displaystyle \int sin^2(x).cos(2x) dx\)
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.....
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.....