Calc1.63

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$\int \sin^2x\cos^2x\,dx\,$

When both $sin x$ and $cos x$ have even powers, turn both into $cos(2 x)$ by using the power reducing formulas

$\sin^2u=\frac{1-\cos (2u)}{2}\,$

$\cos^2u=\frac{1+\cos (2u)}{2}\,$

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

Now, if there is an odd number of $cos x$ , simply turn all but one into $sin x$ and we know the procedure. If there is still an even number of $cos x$ (and there is always an even number of $sin x$ , that is 0), continue to reduce them until you get to an odd power of $cos x$ . First, for to lessen the mess, I will change the $cos$ to $sin$ .

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

Thus

$\int \sin^2x\cos^2x\,dx=\frac{1}{8}x-\frac{1}{32}\sin (4x)+C\,$

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Calc1.63

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