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

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

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Answer 1 · 2016-10-22T14:05:56

$\int {sin}^{2} x {cos}^{2} x d x = \frac{x}{8} - \frac{sin 4 x}{32}$ $intsin^2xcos^2xdx=x/8-(sin4x)/32$

Explanation:

As $sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$ , $sin x cos x = \frac{1}{2} sin 2 x$ $sinxcosx=1/2sin2x$ and

${sin}^{2} x {cos}^{2} x = \frac{1}{4} {sin}^{2} 2 x = \frac{1}{4} (\frac{1}{2} (1 - cos 4 x)) = \frac{1 - cos 4 x}{8}$ $sin^2xcos^2x=1/4sin^2 2x=1/4(1/2(1-cos4x))=(1-cos4x)/8$

Hence $\int {sin}^{2} x {cos}^{2} x d x = \int \frac{1 - cos 4 x}{8} d x$ $intsin^2xcos^2xdx=int(1-cos4x)/8 dx$

= $\frac{x}{8} - \frac{1}{8} \int cos 4 x d x$ $x/8-1/8intcos4xdx$

Let $u = 4 x$ $u=4x$ , then $d u = 4 d x$ $du=4dx$

and $\int cos 4 x d x = \frac{1}{4} \int cos u d u = \frac{1}{4} sin u = \frac{1}{4} sin 4 x$ $intcos4xdx=1/4intcosudu=1/4sinu=1/4sin4x$

Hence $\int {sin}^{2} x {cos}^{2} x d x = \frac{x}{8} - \frac{1}{8} \times \frac{1}{4} sin 4 x = \frac{x}{8} - \frac{sin 4 x}{32}$ $intsin^2xcos^2xdx=x/8-1/8xx1/4sin4x=x/8-(sin4x)/32$

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

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