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Chapter 5: Q. 72 (page 452)

Solve each of the definite integrals in Exercises 67鈥�76.
${鈭�}_{- 蟺}^{蟺} \sin^{2} x \cos^{2} x d x$ .

Short Answer

Expert verified

The answer is $\frac{蟺}{4}$ .

Step by step solution

01

Step 1. Given Information.

The integral is ${鈭�}_{- 蟺}^{蟺} \sin^{2} x \cos^{2} x d x$ .

02

Step 2. Explanation.

To simplify the integral use the following identities,

$\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)$ and $\cos^{2} x = \frac{1}{2} (1 + \cos 2 x)$

Integral becomes as follows,

${鈭�}_{- 蟺}^{蟺} \frac{1}{2} (1 - \cos 2 x) \frac{1}{2} (1 + \cos 2 x) d x = \frac{1}{4} {鈭�}_{- 蟺}^{蟺} (1 - \cos 2 x) (1 + \cos 2 x) d x = \frac{1}{4} {鈭�}_{- 蟺}^{蟺} 1 - \cos^{2} 2 x d x$

03

Step 3. Calculation

Further simplify as follows,

$\frac{1}{4} {鈭�}_{- 蟺}^{蟺} 1 - \cos^{2} 2 x d x = \frac{1}{4} {鈭�}_{- 蟺}^{蟺} 1 - (\frac{1}{2} (1 + \cos 4 x)) d x = \frac{1}{4} {鈭�}_{- 蟺}^{蟺} 1 - \frac{1}{2} - + \cos 4 x d x = \frac{1}{8} {鈭�}_{- 蟺}^{蟺} 1 - \cos 4 x d x$

So,

$\frac{1}{8} {鈭�}_{- 蟺}^{蟺} 1 d x - {鈭�}_{- 蟺}^{蟺} \cos 4 x d x = \frac{1}{8} ({[x]}_{- 蟺}^{蟺} - {[\sin 4 x]}_{- 蟺}^{蟺}) = \frac{1}{8} (2 蟺 - 0) = \frac{蟺}{4}$

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