SATHEE: Indefinite Integration 1 Question 20

Indefinite Integration 1 Question 20

21. The integral $\int_{0}^{π} \sqrt{1 + 4 \sin^{2} \frac{x}{2} - 4 \sin \frac{x}{2}} d x$ is equal to

(a) $π - 4$

(b) $\frac{2 π}{3} - 4 - 4 \sqrt{3}$

(d) $4 \sqrt{3} - 4 - π / 3$

(2014 Main)

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Solution:

PLAN Use the formula, $| x - a | = \begin{array}{cc} x - a, & x \geq a - (x - a), & x < a \end{array}$

to break given integral in two parts and then integrate separately.

$\begin{aligned} \int_{0}^{π} \sqrt{1 - 2 \sin \frac{x^{2}}{2}} d x = \int_{0}^{π} | 1 - 2 \sin \frac{x}{2} | d x \\ = \int_{0}^{\frac{π}{3}} 1 - 2 \sin \frac{x}{2} d x - \int_{\frac{π}{3}}^{π} 1 - 2 \sin \frac{x}{2} d x \\ = x + 4 \cos \frac{x}{2}_{0}^{\frac{π}{3}} - x + 4 \cos \frac{x}{2} \frac{π}{3} \\ = 4 \sqrt{3} - 4 - \frac{π}{3} \end{aligned}$

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Indefinite Integration 1 Question 20

21. The integral ∫0π1+4sin2⁡x2−4sin⁡x2dx is equal to

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21. The integral $\int_{0}^{π} \sqrt{1 + 4 \sin^{2} \frac{x}{2} - 4 \sin \frac{x}{2}} d x$ is equal to