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Chapter 9: Q. 16. (page 756)

complete Example 4 by evaluating the integral
${鈭�}_{0}^{2 蟺} \sqrt{2 + 2 \cos 胃} d 胃$

Short Answer

Expert verified

The required answer is $8$

Step by step solution

01

Given information

The integral is ${鈭�}_{0}^{2 蟺} \sqrt{2 + 2 \cos 胃} d 胃$

02

The objective is to find the value of the integral.

The integral ${鈭�}_{0}^{2 蟺} \sqrt{2 + 2 \cos 胃} d 胃 .$

${鈭�}_{0}^{2 蟺} \sqrt{2 + 2 \cos 胃 d 胃} = {鈭�}_{0}^{2 蟺} \sqrt{2 (1 + \cos 胃)} d 胃 = \sqrt{2} {鈭�}_{0}^{2 蟺} \sqrt{(1 + \cos 胃)} d 胃$

Then,

${鈭�}_{0}^{2 蟺} \sqrt{2 + 2 \cos 胃} d 胃 = \sqrt{2} {鈭�}_{0}^{2 蟺} \sqrt{(1 + 2 \cos^{2} \frac{胃}{2} - 1)} d 胃 [\sin c e \cos 胃 = 2 \cos^{2} \frac{胃}{2} - 1] = \sqrt{2} {鈭�}_{0}^{2 蟺} \sqrt{2 \cos^{2} \frac{胃}{2}} d 胃 = \sqrt{2} {鈭�}_{0}^{2 蟺} \sqrt{2} \cos \frac{胃}{2} d 胃$

03

Find the value of integral

In this case, the supplied function has a negative value in the interval $蟺$ to $2 蟺$

Calculate the integral in the range of $0$ to $蟺$ and multiply it by $2$

${鈭�}_{0}^{2 蟺} \sqrt{2 + 2 \cos 胃} d 胃 = 2 [2 {(2 \sin \frac{胃}{2})}_{0}^{蟺}] = 8 (\sin \frac{蟺}{2} - 0) {鈭�}_{0}^{2 蟺} \sqrt{2 + 2 \cos 胃} d 胃 = 8$

Hence, the value of the integral is $8$

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