How do you find the integral of ${sin}^{2} (x) {cos}^{2} (x) d x$ $sin^2 (x)cos^2 (x) dx$ ?

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Answer 1 · 2016-07-19T18:01:08

$\frac{1}{32} (4 x - sin 4 x) + C$ $1/32(4x-sin4x)+C$ .

Let us note that, ${sin}^{2} x {cos}^{2} x = \frac{1}{4} {(sin 2 x)}^{2} = \frac{{sin}^{2} (2 x)}{4}$ $sin^2xcos^2x=1/4(sin2x)^2=sin^2(2x)/4$

Recall that, ${sin}^{2} θ = \frac{1 - cos 2 θ}{2}$ $sin^2theta=(1-cos2theta)/2$

Letting $θ = 2 x$ $theta=2x$ , we have, ${sin}^{2} (2 x) = \frac{1 - cos 4 x}{2}$ $sin^2(2x)=(1-cos4x)/2$

$:. intsin^2xcos^2xdx=1/8int(1-cos4x)dx=1/8(x-sin(4x)/4)=1/32(4x-sin4x)+C$ .

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