The integral ∫ 0 π 1 + 4 sin 2 x 2 - 4 sin x 2 d x equals

∫ 0 π 1 + 4 sin 2 x 2 - 4 sin x 2 d x = ∫ 0 π 2 sin x 2 - 1 2 d x = ∫ 0 π 2 sin x 2 - 1 d x = 2 ∫ 0 π sin x 2 - 1 2 d x = 2 ∫ 0 π / 3 sin x 2 - 1 2 d x + ∫ π / 3 π sin x 2 - 1 2 d x = 2 ∫ 0 π / 3 - sin x 2 - 1 2 d x + ∫ π / 3 π sin x 2 - 1 2 d x = 2 2 cos x 2 + x 2 0 π / 3 + - 2 cos x 2 - x 2 π / 3 π = 2 2 3 2 + π 6 - 2 + - π 2 + 2 3 2 + π 6 = 2 3 + π 6 - 2 - π 2 + 3 + π 6 = 2 2 3 - π 6 - 2 = 4 3 - π 3 - 4 .

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Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationProperties of definite integrationMedium2 minPYQ_2014

MathematicsMediumsingle choice

The integral $\int_{0}^{π} \sqrt{1 + 4 {sin}^{2} \frac{x}{2} - 4 sin \frac{x}{2}} d x$ equals

Question diagram: The integral ∫ 0 π 1 + 4 sin 2 x 2 - 4 sin x 2 d x equals

Options:

Answer:

Solution:

$\int_{0}^{π} \sqrt{1 + 4 {sin}^{2} \frac{x}{2} - 4 sin \frac{x}{2}} d x$

$= \int_{0}^{π} \sqrt{{(2 sin \frac{x}{2} - 1)}^{2}} d x$

$= \int_{0}^{π} |2 sin \frac{x}{2} - 1| d x$

$= 2 \int_{0}^{π} |sin \frac{x}{2} - \frac{1}{2}| d x$

$= 2 [\int_{0}^{π / 3} |sin \frac{x}{2} - \frac{1}{2}| d x + \int_{π / 3}^{π} |sin \frac{x}{2} - \frac{1}{2}| d x]$

$= 2 [\int_{0}^{π / 3} - (sin \frac{x}{2} - \frac{1}{2}) d x + \int_{π / 3}^{π} (sin \frac{x}{2} - \frac{1}{2}) d x]$

$= 2 [{[2 cos \frac{x}{2} + \frac{x}{2}]}_{0}^{π / 3} + {[- 2 cos \frac{x}{2} - \frac{x}{2}]}_{π / 3}^{π}]$

$= 2 [[2 (\frac{\sqrt{3}}{2}) + \frac{π}{6} - 2] + [- \frac{π}{2} + 2 (\frac{\sqrt{3}}{2}) + \frac{π}{6}]]$

$= 2 [\sqrt{3} + \frac{π}{6} - 2 - \frac{π}{2} + \sqrt{3} + \frac{π}{6}]$

$= 2 [2 \sqrt{3} - \frac{π}{6} - 2]$

$= 4 \sqrt{3} - \frac{π}{3} - 4$ .

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Properties of definite integration

⏱ 2mℹ️ Source: PYQ_2014

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