How do you find the integral of ${cos (x)}^{2} \cdot {sin (x)}^{2}$ $cos(x)^2*sin(x)^2$ ?

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How do you find the integral of ${cos (x)}^{2} \cdot {sin (x)}^{2}$ $cos(x)^2*sin(x)^2$ ?

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Answer 1 · 2016-10-21T01:11:16

$\frac{x}{8} - \frac{1}{32} sin (4 x) + C$ $x/8-1/32sin(4x)+C$

Explanation:

This can be written as:

$I = \int {cos}^{2} (x) {sin}^{2} (x) d x$ $I=intcos^2(x)sin^2(x)dx$

We can rewrite this using the identity $sin (2 x) = 2 sin (x) cos (x)$ $sin(2x)=2sin(x)cos(x)$ . Thus, $sin (x) cos (x) = \frac{sin (2 x)}{2}$ $sin(x)cos(x)=sin(2x)/2$ . Square both sides to see that ${cos}^{2} (x) {sin}^{2} (x) = \frac{{sin}^{2} (2 x)}{4}$ $cos^2(x)sin^2(x)=sin^2(2x)/4$ . Thus:

$I = \frac{1}{4} \int {sin}^{2} (2 x) d x$ $I=1/4intsin^2(2x)dx$

Let $u = 2 x$ $u=2x$ so that $d u = 2 d x$ $du=2dx$ .

$I = \frac{1}{8} \int {sin}^{2} (2 x) \cdot (2 d x) = \frac{1}{8} \int {sin}^{2} (u) d u$ $I=1/8intsin^2(2x)*(2dx)=1/8intsin^2(u)du$

The way to integrate this is to use another double-angle formula.

$cos (2 u) = 1 - 2 {sin}^{2} (u) \Rightarrow {sin}^{2} (u) = \frac{1}{2} (1 - cos (2 u))$ $cos(2u)=1-2sin^2(u)" "=>" "sin^2(u)=1/2(1-cos(2u))$

Thus:

$I = \frac{1}{16} \int (1 - cos (2 u)) d u$ $I=1/16int(1-cos(2u))du$

$I = \frac{1}{16} \int d u - \frac{1}{16} \int cos (2 u) d u$ $I=1/16intdu-1/16intcos(2u)du$

The first integral is the most basic integral and the second can be solved through inspection or substitution, where $v = 2 u$ $v=2u$ .

$I = \frac{1}{16} u - \frac{1}{32} sin (2 u) + C$ $I=1/16u-1/32sin(2u)+C$

Since $u = 2 x$ $u=2x$ :

$I = \frac{x}{8} - \frac{1}{32} sin (4 x) + C$ $I=x/8-1/32sin(4x)+C$

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How do you find the integral of ${cos (x)}^{2} \cdot {sin (x)}^{2}$ $cos(x)^2*sin(x)^2$ ?

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