Question

integerate sin^2x cos^2x

Answer 1

\[\int\limits_{0}^{\pi/2} \sin ^{2}x*\cos ^{2}x \]

Answer 2

\[(\sin ^{2 }x +\cos ^{2}x)^{2}=(\sin ^{4}x)+(\cos ^{4}x) +2 \sin ^{2}x \cos ^{2}x\]
\[2\sin ^{2}x \cos ^{2}x=1-\sin ^{4}(x)-\cos ^{4}(x)\] since

Answer 3

\[\sin ^{2}x+\cos ^{2}x=1\]

Answer 4

Yu can use the direct formula for finding \[\int\limits_{0}^{\pi/2}\sin ^{4}(x) dx\]

Answer 5

so sin^2x * cos^ 2x =1-sin^4x - cos^4x/2

Answer 6

yups...

Answer 7

Or you could do the other way: \[\sin(2x)=2 \sin(x) \cos(x)\]
\[\sin ^{2}(2x)=4\sin ^{2}(x)\cos ^{2}(x)\] \[(1-\cos(4x))/2= \sin ^{2}(2x)\]

Answer 8

\[\int\limits_{0}^{\pi/2}((1-\cos(4x))/8) dx\]

Answer 9

This will be your reduced integral in this case..

Answer 10

thanks a lot nitz.

Answer 11

You are welcome!!